理論電磁気学の行間埋め　第12章

\begin{eqnarray} \delta I &=& \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q(t)}\delta q(t)+\frac{\partial L}{\partial \dot{q}(t)}\frac{d}{dt}\delta q(t)\right)dt \\ \\ &=& \int_{t_1}^{t_2}\frac{\partial L}{\partial q(t)}\delta q(t)dt+\int_{t_1}^{t_2}\frac{\partial L}{\partial \dot{q}(t)}\frac{d}{dt}\delta q(t)dt \\ \\ &=& \int_{t_1}^{t_2}\frac{\partial L}{\partial q(t)}\delta q(t)dt+\left[\frac{\partial L}{\partial \dot{q}(t)}\delta q(t)\right]_{t_1}^{t_2}-\int_{t_1}^{t_2}\frac{d}{dt}\frac{\partial L}{\partial \dot{q}(t)}\delta q(t)dt \\ \\ &=& \int_{t_1}^{t_2}\frac{\partial L}{\partial q(t)}\delta q(t)dt-\int_{t_1}^{t_2}\frac{d}{dt}\frac{\partial L}{\partial \dot{q}(t)}\delta q(t)dt+\left[\frac{\partial L}{\partial \dot{q}(t_1)}\delta q(t_1)-\frac{\partial L}{\partial \dot{q}(t_2)}\delta q(t_2)\right] \\ \\ &=& \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q(t)}\delta q(t)-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}(t)}\delta q(t)\right)dt+\left[\frac{\partial L}{\partial \dot{q}(t_1)}0-\frac{\partial L}{\partial \dot{q}(t_2)}0\right]&...&\text{式(1.4)より} \\ \\ &=& \int_{t_1}^{t_2}\left(\frac{\partial L}{\partial q(t)}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}(t)}\right)\delta q(t)dt& \\ \\ \end{eqnarray} が得られる。

\begin{eqnarray} \delta\int_VL_Fd^4x &=& -\frac{\varepsilon_0}{2}\int_Vd^4x\left(F_{\mu\nu}(x)\frac{\partial}{\partial x_{\mu}}\delta A_{\nu}(x)-F_{\mu\nu}(x)\frac{\partial}{\partial x_{\nu}}\delta A_{\mu}(x)\right) \\ \\ &=& -\frac{\varepsilon_0}{2}\int_Vd^4x\left(-\frac{\partial F_{\mu\nu}(x)}{\partial x_{\mu}}\delta A_{\nu}(x)-F_{\mu\nu}(x)\frac{\partial}{\partial x_{\nu}}\delta A_{\mu}(x)\right)&...&(1) \\ \\ &=& -\frac{\varepsilon_0}{2}\int_Vd^4x\left(-\frac{\partial F_{\mu\nu}(x)}{\partial x_{\mu}}\delta A_{\nu}(x)+\frac{\partial F_{\mu\nu}(x)}{\partial x_{\nu}}\delta A_{\mu}(x)\right)&...&(2) \\ \\ &=& \frac{\varepsilon_0}{2}\int_Vd^4x\left(\frac{\partial F_{\mu\nu}(x)}{\partial x_{\mu}}\delta A_{\nu}(x)-\frac{\partial F_{\mu\nu}(x)}{\partial x_{\nu}}\delta A_{\mu}(x)\right)& \\ \\ &=& \frac{\varepsilon_0}{2}\int_Vd^4x\left(\frac{\partial F_{\nu\mu}(x)}{\partial x_{\nu}}\delta A_{\mu}(x)-\frac{\partial F_{\mu\nu}(x)}{\partial x_{\nu}}\delta A_{\mu}(x)\right)&...&\text{総和の記法ルールに従って、変数}\mu,\nu\text{を入れ替えた。} \\ \\ &=& \frac{\varepsilon_0}{2}\int_Vd^4x\left(\frac{\partial F_{\nu\mu}(x)}{\partial x_{\nu}}-\frac{\partial F_{\mu\nu}(x)}{\partial x_{\nu}}\right)\delta A_{\mu}(x)& \\ \\ &=& \frac{\varepsilon_0}{2}\int_Vd^4x\left(-\frac{\partial F_{\mu\nu}(x)}{\partial x_{\nu}}-\frac{\partial F_{\mu\nu}(x)}{\partial x_{\nu}}\right)\delta A_{\mu}(x)&...&F_{\mu\nu}=-F_{\nu\mu}\text{を利用(p.381など)。} \\ \\ &=& -\varepsilon_0\int_Vd^4x\frac{\partial F_{\mu\nu}(x)}{\partial x_{\nu}}\delta A_{\mu}(x)& \\ \\ \end{eqnarray} が得られる。

1. (1)部分積分を利用(i)
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成分\(\mu=i\)に着目する。 \begin{eqnarray} &&-\frac{\varepsilon_0}{2}\int_Vd^4xF_{\mu\nu}(x)\frac{\partial}{\partial x_{\mu}}\delta A_{\nu}(x) \\ \\ &\Rightarrow& -\frac{\varepsilon_0}{2}\int_Vd^4x\left[F_{i\nu}(x)\frac{\partial}{\partial x_{i}}\delta A_{\nu}(x)\right] \\ \\ &=& -\frac{\varepsilon_0}{2}\left[F_{i\nu}(x)\delta A_{\nu}(x)\right]_V+\frac{\varepsilon_0}{2}\int_Vd^4x\left[\frac{\partial F_{i\nu}(x)}{\partial x_{i}}\delta A_{\nu}(x)\right] \\ \\ \end{eqnarray} が得られる。ここで、\(\delta A_{mu}(x)\)は\(V\)の無限遠方で0になると仮定する(p.417下)ので \begin{eqnarray} &&-\frac{\varepsilon_0}{2}\left[F_{i\nu}(x)\delta A_{\nu}(x)\right]_V+\frac{\varepsilon_0}{2}\int_Vd^4x\left[\frac{\partial F_{i\nu}(x)}{\partial x_{i}}\delta A_{\nu}(x)\right] \\ \\ &=& \frac{\varepsilon_0}{2}\int_Vd^4x\left[\frac{\partial F_{i\nu}(x)}{\partial x_{i}}\delta A_{\nu}(x)\right] \\ \\ \end{eqnarray} が得られ、これはすべての\(\nu\)に対して同様なので、 \begin{eqnarray} -\frac{\varepsilon_0}{2}\int_Vd^4xF_{\mu\nu}(x)\frac{\partial}{\partial x_{\mu}}\delta A_{\nu}(x) &=& \frac{\varepsilon_0}{2}\int_Vd^4x\frac{\partial F_{\mu\nu}(x)}{\partial x_{\mu}}\delta A_{\nu}(x) \\ \\ \end{eqnarray} となる。


2. (2)部分積分を利用(ii)
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(i)と同様に成分\(\nu=i\)に着目する。 \begin{eqnarray} \frac{\varepsilon_0}{2}\int_Vd^4xF_{\mu\nu}(x)\frac{\partial}{\partial x_{\nu}}\delta A_{\mu}(x) &\Rightarrow& \frac{\varepsilon_0}{2}\int_Vd^4x\left[F_{\mu i}(x)\frac{\partial}{\partial x_{i}}\delta A_{\mu}(x)\right] \\ \\ &=& -\frac{\varepsilon_0}{2}\int_Vd^4x\left[F_{i\mu }(x)\frac{\partial}{\partial x_{i}}\delta A_{\mu}(x)\right] \\ \\ \end{eqnarray} が得られる。これは先ほどの部分積分にマイナスが付いたものなので、同様の仮定を置くことで、 \begin{eqnarray} &&\frac{\varepsilon_0}{2}\int_Vd^4xF_{\mu\nu}(x)\frac{\partial}{\partial x_{\nu}}\delta A_{\mu}(x) \\ \\ &=& -\frac{\varepsilon_0}{2}\int_Vd^4xF_{\nu\mu}(x)\frac{\partial}{\partial x_{\nu}}\delta A_{\mu}(x)&...&\text{反対称であることを利用} \\ \\ &=& -\frac{\varepsilon_0}{2}\int_Vd^4xF_{\mu\nu}(x)\frac{\partial}{\partial x_{\mu}}\delta A_{\nu}(x)&...&\nu,\mu\text{を入れ替えて(i)の式に置換} \\ \\ &=& \frac{\varepsilon_0}{2}\int_Vd^4x\frac{\partial F_{\mu\nu}(x)}{\partial x_{\mu}}\delta A_{\nu}(x) \\ \\ &=& -\frac{\varepsilon_0}{2}\int_Vd^4x\frac{\partial F_{\nu\mu}(x)}{\partial x_{\mu}}\delta A_{\nu}(x) \\ \\ &=& -\frac{\varepsilon_0}{2}\int_Vd^4x\frac{\partial F_{\mu\nu}(x)}{\partial x_{\nu}}\delta A_{\mu}(x)&...&\nu,\mu\text{を入れ替えて(ii)の式に置換} \\ \\ \\ \\ \end{eqnarray} となる。

p.417上に、はじめに電磁場\(A_{\mu}(x)\)について変分をとる、とあるので、変分はその項のみに作用させている。

\begin{eqnarray} && e_i\int_{-\infty}^{\infty}d\tau\delta^4(x-z^i(\tau))\frac{dz_{\mu}^i(\tau)}{d\tau} \\ \\ &=& e_i\int_{-\infty}^{\infty}d\tau\delta^3(\boldsymbol{x}-\boldsymbol{z}^i(\tau))\delta(t-t(\tau))\frac{dz_{\mu}^i(\tau)}{d\tau}&...&\text{式(2.6)より} \\ \\ &=& e_i\int_{-\infty}^{\infty}\left(\frac{d\tau}{\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}}\right)\delta^3(\boldsymbol{x}-\boldsymbol{z}^i(\tau))\delta(t-t(\tau))\frac{dz_{\mu}^i(\tau)\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}}{d\tau}& \\ \\ &=& e_i\int_{-\infty}^{\infty}dt'\delta^3(\boldsymbol{x}-\boldsymbol{z}^i(t'))\delta(t-t')\frac{dz_{\mu}^i(\tau)}{dt'}&...&\text{p.394下より}dt'=\frac{d\tau}{\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}} \\ \\ &=& e_i\delta^3(\boldsymbol{x}-\boldsymbol{z}^i(t))\frac{dz_{\mu}^i(\tau)}{dt}&...&\text{デルタ関数の積分より}t'=t\text{の項が残る。p.453下} \\ \\ &=& \boldsymbol{i}^i(\boldsymbol{x},t)&...&\text{p.29式(1.4)など} \\ \\ \end{eqnarray} と導出できる。

\begin{eqnarray} && e_i\int_{-\infty}^{\infty}d\tau\delta^4(x-z^i(\tau))\frac{dz_{0}^i(\tau)}{d\tau} \\ \\ &=& e_i\int_{-\infty}^{\infty}d\tau\delta^3(\boldsymbol{x}-\boldsymbol{z}^i(\tau))\delta(t-t(\tau))\frac{dz_{0}^i(\tau)}{d\tau}&...&\text{式(2.6)より} \\ \\ &=& e_i\int_{-\infty}^{\infty}\left(\frac{d\tau}{\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}}\right)\delta^3(\boldsymbol{x}-\boldsymbol{z}^i(\tau))\delta(t-t(\tau))\frac{dz_{0}^i(\tau)\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}}{d\tau}& \\ \\ &=& e_i\int_{-\infty}^{\infty}\left(\frac{d\tau}{\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}}\right)\delta^3(\boldsymbol{x}-\boldsymbol{z}^i(\tau))\delta(t-t(\tau))w_0(\tau)\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}&...&\text{式(4.9)より} \\ \\ &=& e_i\int_{-\infty}^{\infty}\left(\frac{d\tau}{\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}}\right)\delta^3(\boldsymbol{x}-\boldsymbol{z}^i(\tau))\delta(t-t(\tau))\frac{c}{\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}}\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}&...&\text{式(4.11)より} \\ \\ &=& e_i\int_{-\infty}^{\infty}dt'\delta^3(\boldsymbol{x}-\boldsymbol{z}^i(t'))\delta(t-t')&...&\text{p.394下より}dt'=\frac{d\tau }{\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}}\\ \\ &=& e_i\delta^3(\boldsymbol{x}-\boldsymbol{z}^i(t))c&...&\text{デルタ関数の積分より}t'=t\text{の項が残る。p.453下} \\ \\ &=& c\rho^i(\boldsymbol{x},t)&...&\text{p.29式(1.3)など} \\ \\ \end{eqnarray} と導出できる。

\begin{eqnarray} \text{式(2.16)左辺}&=& -\delta\int_{-\infty}^{\infty}mc^2d\tau+\delta\int_{-\infty}^{\infty}\frac{e}{c}A_{\mu}(z(\tau))\frac{dz_{\mu}(\tau)}{d\tau}d\tau \\ \\ &=& -\delta\int_{-\infty}^{\infty}mc^2d\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}dt+\delta\int_{-\infty}^{\infty}\frac{e}{c}A_{\mu}(z(\tau))\frac{dz_{\mu}(\tau)}{d\tau}d\tau&...&\text{p.394下式より} \\ \\ &=& -\delta\int_{-\infty}^{\infty}mc^2d\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}dt+\delta\int_{-\infty}^{\infty}\frac{e}{c}A_{\mu}(z(\tau))w_{\mu}(t(\tau))d\tau&...&\text{p.394式(4.9)より} \\ \\ &=& -\delta\int_{-\infty}^{\infty}mc^2d\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}dt+\delta\int_{-\infty}^{\infty}\frac{e}{c}A_{\mu}(z(t))w_{\mu}(t)\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}dt&...&\text{p.394下式より} \\ \\ &=& -\delta\int_{-\infty}^{\infty}mc^2d\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}dt+\delta\int_{-\infty}^{\infty}\frac{e}{c}\left(A_{1}(z(t))w_{1}(t)+A_{2}(z(t))w_{2}(t)+A_{3}(z(t))w_{3}(t)+A_{4}(z(t))w_{4}(t)\right)\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}dt& \\ \\ &=& -\delta\int_{-\infty}^{\infty}mc^2d\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}dt+\delta\int_{-\infty}^{\infty}\frac{e}{c}\left(A_{1}(z(t))\frac{u_{x}(t)}{\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}}+A_{2}(z(t))\frac{u_{y}(t)}{\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}}+A_{3}(z(t))\frac{u_{z}(t)}{\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}}+A_{4}(z(t))\frac{ic}{\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}}\right)\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}dt&...&\text{p.395式(4.10)(4.11)}iw_0=w_4\text{より} \\ \\ &=& -\delta\int_{-\infty}^{\infty}mc^2d\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}dt+\delta\int_{-\infty}^{\infty}\frac{e}{c}\left(A_{1}(z(t))u_{x}(t)+A_{2}(z(t))u_{y}(t)+A_{3}(z(t))u_{z}(t)+A_{4}(z(t))ic\right)dt& \\ \\ &=& -\delta\int_{-\infty}^{\infty}mc^2d\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}dt+\delta\int_{-\infty}^{\infty}\frac{e}{c}\left(cA_{x}(z(t))u_{1}(t)+cA_{y}(z(t))u_{2}(t)+cA_{z}(z(t))u_{3}(t)+(iA_{0}(z(t)))ic\right)dt&...&\text{p.385式(3.43)より} \\ \\ &=& -\delta\int_{-\infty}^{\infty}mc^2d\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}dt+\delta\int_{-\infty}^{\infty}e\left(A_{x}(z(t))u_{1}(t)+A_{y}(z(t))u_{2}(t)+A_{z}(z(t))u_{3}(t)-A_{0}(z(t))\right)dt&\\ \\ &=& -\delta\int_{-\infty}^{\infty}mc^2d\sqrt{1-\frac{\boldsymbol{u}^2}{c^2}}dt+\delta\int_{-\infty}^{\infty}e\left(\boldsymbol{A}\cdot\boldsymbol{u}-A_{0}(z(t))\right)dt&\\ \\ \end{eqnarray} と導出できる。

\begin{eqnarray} &&-\delta\int_{-\infty}^{\infty}mc^2\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}dt \\ \\ &=& - mc^2\int_{-\infty}^{\infty}\delta\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}dt \\ \\ &=& - mc^2\int_{-\infty}^{\infty}\frac{-\boldsymbol{u}(t)\cdot\delta\boldsymbol{u}(t)}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}}dt \\ \\ &=& mc^2\int_{-\infty}^{\infty}\frac{\boldsymbol{u}(t)\cdot\delta\left(\frac{d\boldsymbol{z}(t)}{dt}\right)}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}}dt&...&\text{p.418中段より} \\ \\ &=& mc^2\int_{-\infty}^{\infty}\frac{\boldsymbol{u}(t)}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}}\cdot\frac{d}{dt}\delta\boldsymbol{z}(t)dt&...&\text{微分と変分は入れ替えられるため} \\ \\ &=& -mc^2\int_{-\infty}^{\infty}\frac{d}{dt}\left(\frac{\boldsymbol{u}(t)}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}}\right)\cdot\delta\boldsymbol{z}(t)dt&...&(1) \\ \\ \end{eqnarray} と導出できる。

1. (1)部分積分を利用
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\begin{eqnarray} &&mc^2\int_{-\infty}^{\infty}\frac{\boldsymbol{u}(t)}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}}\cdot\frac{d}{dt}\delta\boldsymbol{z}(t)dt& \\ \\ &=& mc^2\left[\frac{\boldsymbol{u}(t)}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}}\cdot\delta\boldsymbol{z}(t)\right]_{-\infty}^{\infty}-mc^2\int_{-\infty}^{\infty}\frac{d}{dt}\left(\frac{\boldsymbol{u}(t)}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}}\right)\cdot\delta\boldsymbol{z}(t)dt& \\ \\ &=& -mc^2\int_{-\infty}^{\infty}\frac{d}{dt}\left(\frac{\boldsymbol{u}(t)}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}}\right)\cdot\delta\boldsymbol{z}(t)dt& \\ \\ \end{eqnarray} と変形できる。ここで、\(\delta z(t)\)の\(t\to\infty,t\to-\infty\)において、運動の端点では変分が0になるとした。式(1.4)などを参考。
もっと良い説明があると思う。

\begin{eqnarray} &&\delta\int_{-\infty}^{\infty}e\left(\boldsymbol{A}(\boldsymbol{z}(t),t)\cdot\boldsymbol{u}(t)-A_0(\boldsymbol{z}(t),t)\right)dt \\ \\ &=& e\int_{-\infty}^{\infty}\left(\delta\boldsymbol{A}(\boldsymbol{z}(t),t)\cdot\boldsymbol{u}(t)+\boldsymbol{A}(\boldsymbol{z}(t),t)\cdot\delta\boldsymbol{u}(t)-\delta A_0(\boldsymbol{z}(t),t)\right)dt \\ \\ &=& e\int_{-\infty}^{\infty}\left[\displaystyle\sum_k\delta A_k(\boldsymbol{z}(t),t) u_k(t)+\sum_kA_k(\boldsymbol{z}(t),t)\delta u_k(t)-\delta A_0(\boldsymbol{z}(t),t)\right]dt \\ \\ &=& e\int_{-\infty}^{\infty}\left[\displaystyle\sum_{i,k}\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}\delta z_i(t) u_k(t)+\sum_kA_k(\boldsymbol{z}(t),t)\delta u_k(t)-\delta A_0(\boldsymbol{z}(t),t)\right]dt&...&(1) \\ \\ &=& e\int_{-\infty}^{\infty}\left[\displaystyle\sum_{i,k}\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}\delta z_i(t) u_k(t)+\sum_kA_k(\boldsymbol{z}(t),t)\delta \frac{dz_k(t)}{dt}-\delta A_0(\boldsymbol{z}(t),t)\right]dt&...&\text{p.418中段より} \\ \\ &=& e\int_{-\infty}^{\infty}\left[\displaystyle\sum_{i,k}\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}\delta z_i(t) u_k(t)+\sum_kA_k(\boldsymbol{z}(t),t) \frac{d\delta z_k(t)}{dt}- \sum_i\frac{\partial A_0(\boldsymbol{z}(t),t)}{\partial z_i}\delta z_i(t)\right]dt&...&(1),\text{変分と微分の入れ替え} \\ \\ &=& e\int_{-\infty}^{\infty}\left[\displaystyle\sum_{i,k}\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}\delta z_i(t) u_k(t)-\sum_k\frac{dA_k(\boldsymbol{z}(t),t)}{dt} \delta z_k(t)- \sum_i\frac{\partial A_0(\boldsymbol{z}(t),t)}{\partial z_i}\delta z_i(t)\right]dt&...&(2) \\ \\ &=& e\int_{-\infty}^{\infty}\left[\displaystyle\sum_{i,k}\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}\delta z_i(t) u_k(t)-\sum_{i,k}\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}\frac{\partial z_i(t)}{\partial t} \delta z_k(t)-\sum_{i}\frac{\partial A_i}{\partial t}\delta z_i(t) -\sum_i\frac{\partial A_0(\boldsymbol{z}(t),t)}{\partial z_i}\delta z_i(t)\right]dt&...&(3) \\ \\ &=& e\int_{-\infty}^{\infty}\left[\displaystyle\sum_{i,k}\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}\delta z_i(t) u_k(t)-\sum_{i,k}\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}u_i(t) \delta z_k(t)-\sum_{i}\frac{\partial A_i}{\partial t}\delta z_i(t) -\sum_i\frac{\partial A_0(\boldsymbol{z}(t),t)}{\partial z_i}\delta z_i(t)\right]dt&...&\text{p.418中段より} \\ \\ &=& e\int_{-\infty}^{\infty}\displaystyle\sum_i\left[\left(-\frac{\partial A_0(\boldsymbol{z}(t),t)}{\partial z_i}-\frac{\partial A_i}{\partial t}\right)+\displaystyle\sum_k\left(\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}u_k(t)-\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}u_i(t)\right)\right]dt&\\ \\ &=& e\int_{-\infty}^{\infty}\displaystyle\sum_i\left[\left(-\frac{\partial A_0(\boldsymbol{z}(t),t)}{\partial z_i}-\frac{\partial A_i}{\partial t}\right)+\displaystyle\sum_k\left(\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}u_k(t)-\frac{\partial A_i(\boldsymbol{z}(t),t)}{\partial z_k}u_k(t)\right)\right]dt&...&\text{表記の問題で}i,k\text{を入れ替えた。}\\ \\ &=& e\int_{-\infty}^{\infty}\displaystyle\sum_i\left[\left(-\frac{\partial A_0(\boldsymbol{z}(t),t)}{\partial z_i}-\frac{\partial A_i}{\partial t}\right)+\displaystyle\sum_k\left(\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}-\frac{\partial A_i(\boldsymbol{z}(t),t)}{\partial z_k}\right)u_k(t)\right]dt&\\ \\ &=& e\int_{-\infty}^{\infty}\displaystyle\sum_i\left[E_i(z(t),t)+\displaystyle\sum_k\left(\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}-\frac{\partial A_i(\boldsymbol{z}(t),t)}{\partial z_k}\right)u_k(t)\right]dt&...&{p.385式(3.43)とp.46式(3.7)より}\\ \\ &=& e\int_{-\infty}^{\infty}\displaystyle\sum_i\left[E_i(z(t),t)+(\boldsymbol{u}\times\boldsymbol{B}(z(t),t))_i\right]\delta z_i dt&...&(4)\\ \\ &=& e\int_{-\infty}^{\infty}\left[\boldsymbol{E}(z(t),t)+\boldsymbol{u}\times\boldsymbol{B}(z(t),t)\right]\cdot\delta \boldsymbol{z}(t) dt&\\ \\ \end{eqnarray} と導出できる。

1. (1)全微分(変分？)を利用
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\begin{eqnarray} &&\delta A_k(\boldsymbol{z}(t),t) &=& \displaystyle\sum_i\frac{\partial A_k}{\partial z_i}\delta z_i \end{eqnarray} と変形できる。ここでは、p.417の上部より\(z_i\)についての変分に着目している。
\(t\)については全微分(変分？)しなくてよいのかは疑問。


2. (2)部分積分を利用した
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\begin{eqnarray} && \int_{-\infty}^{\infty}\left[-A_k(z(t),t)\frac{d\delta z_k(t)}{dt}\right]dt\\ \\ &=& \left[-A_k(z(t),t)\delta z_k(t)\right]_{-\infty}^{\infty}-\int_{-\infty}^{\infty}\left[-\frac{d A_k(z(t),t)}{dt}\delta z_k(t)\right]dt\\ \\ &=& \int_{-\infty}^{\infty}\left[\frac{d A_k(z(t),t)}{dt}\delta z_k(t)\right]dt\\ \\ \end{eqnarray} と変形できる。ここで、端点では変分が0になることを利用し\(\delta z(t)\to 0(t\to \pm\infty)\)とした。式(1.4)など参考。


3. (3)Lagrangeの微分を利用
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p.316式(1.9)やp.317式(1.12)のような、なまの\(t\)に関する微分だけでなく変数に含まれる\(t\)も微分を表すことを利用する。 \begin{eqnarray} \frac{d}{dt}A_k(z(t),t) &=& \frac{\partial}{\partial t}A_k(z(t),t)+\frac{\partial\boldsymbol{z}(t)}{\partial t}\cdot\nabla A_k(z(t),t) \\ \\ &=& \frac{\partial}{\partial t}A_k(z(t),t)+\frac{\partial z_x(t)}{\partial t}\frac{\partial A_k(z(t),t)}{\partial z_x}+\frac{\partial z_y(t)}{\partial t}\frac{\partial A_k(z(t),t)}{\partial z_y}+\frac{\partial z_z(t)}{\partial t}\frac{\partial A_k(z(t),t)}{\partial z_z} \\ \\ &=& \frac{\partial}{\partial t}A_k(z(t),t)+\displaystyle\sum_i\frac{\partial A_k(z(t),t)}{\partial z_i}\frac{\partial z_i(t)}{\partial t} \\ \\ \end{eqnarray} を適用した。


4. (4)\(\displaystyle\sum_k\left(\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}-\frac{\partial A_i(\boldsymbol{z}(t),t)}{\partial z_k}\right)u_k(t)=\boldsymbol{u}\times\boldsymbol{B}|_i\)を利用
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\begin{eqnarray} &&\displaystyle\sum_k\left(\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z_i}-\frac{\partial A_i(\boldsymbol{z}(t),t)}{\partial z_k}\right)u_k(t) \\ \\ &=& \left( \begin{array}{cccc} \displaystyle\sum_k\left(\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial x}-\frac{\partial A_x(\boldsymbol{z}(t),t)}{\partial z_k}\right)u_k(t) \\ \displaystyle\sum_k\left(\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial y}-\frac{\partial A_y(\boldsymbol{z}(t),t)}{\partial z_k}\right)u_k(t) \\ \displaystyle\sum_k\left(\frac{\partial A_k(\boldsymbol{z}(t),t)}{\partial z}-\frac{\partial A_z(\boldsymbol{z}(t),t)}{\partial z_k}\right)u_k(t) \end{array} \right) \\ \\ &=& \left( \begin{array}{cccc} \left(\frac{\partial A_x(\boldsymbol{z}(t),t)}{\partial x}-\frac{\partial A_x(\boldsymbol{z}(t),t)}{\partial x}\right)u_x(t)+\left(\frac{\partial A_y(\boldsymbol{z}(t),t)}{\partial x}-\frac{\partial A_x(\boldsymbol{z}(t),t)}{\partial y}\right)u_y(t)+\left(\frac{\partial A_z(\boldsymbol{z}(t),t)}{\partial x}-\frac{\partial A_x(\boldsymbol{z}(t),t)}{\partial z}\right)u_z(t) \\ \\ \left(\frac{\partial A_x(\boldsymbol{z}(t),t)}{\partial y}-\frac{\partial A_y(\boldsymbol{z}(t),t)}{\partial x}\right)u_x(t)+\left(\frac{\partial A_y(\boldsymbol{z}(t),t)}{\partial y}-\frac{\partial A_y(\boldsymbol{z}(t),t)}{\partial y}\right)u_y(t)+\left(\frac{\partial A_z(\boldsymbol{z}(t),t)}{\partial y}-\frac{\partial A_y(\boldsymbol{z}(t),t)}{\partial z}\right)u_z(t) \\ \\ \left(\frac{\partial A_x(\boldsymbol{z}(t),t)}{\partial z}-\frac{\partial A_z(\boldsymbol{z}(t),t)}{\partial x}\right)u_x(t)+\left(\frac{\partial A_y(\boldsymbol{z}(t),t)}{\partial z}-\frac{\partial A_z(\boldsymbol{z}(t),t)}{\partial y}\right)u_y(t)+\left(\frac{\partial A_z(\boldsymbol{z}(t),t)}{\partial z}-\frac{\partial A_z(\boldsymbol{z}(t),t)}{\partial z}\right)u_z(t) \\ \\ \end{array} \right) \\ \\ &=& \left( \begin{array}{cccc} \left(\frac{\partial A_y(\boldsymbol{z}(t),t)}{\partial x}-\frac{\partial A_x(\boldsymbol{z}(t),t)}{\partial y}\right)u_y(t)+\left(\frac{\partial A_z(\boldsymbol{z}(t),t)}{\partial x}-\frac{\partial A_x(\boldsymbol{z}(t),t)}{\partial z}\right)u_z(t) \\ \\ \left(\frac{\partial A_x(\boldsymbol{z}(t),t)}{\partial y}-\frac{\partial A_y(\boldsymbol{z}(t),t)}{\partial x}\right)u_x(t)+\left(\frac{\partial A_z(\boldsymbol{z}(t),t)}{\partial y}-\frac{\partial A_y(\boldsymbol{z}(t),t)}{\partial z}\right)u_z(t) \\ \\ \left(\frac{\partial A_x(\boldsymbol{z}(t),t)}{\partial z}-\frac{\partial A_z(\boldsymbol{z}(t),t)}{\partial x}\right)u_x(t)+\left(\frac{\partial A_y(\boldsymbol{z}(t),t)}{\partial z}-\frac{\partial A_z(\boldsymbol{z}(t),t)}{\partial y}\right)u_y(t) \\ \\ \end{array} \right) \\ \\ &=& \left( \begin{array}{cccc} B_z(z(t),t)u_y(t)-B_y(z(t),t)u_z(t) \\ \\ -B_z(z(t),t)u_x(t)+B_x(z(t),t)u_z(t) \\ \\ B_y(z(t),t)u_x(t)-B_x(z(t),t)u_y(t) \\ \\ \end{array} \right) \\ \\ &=& \left( \begin{array}{cccc} u_y(t)B_z(z(t),t)-u_z(t)B_y(z(t),t) \\ \\ u_z(t)B_x(z(t),t)-u_x(t)B_z(z(t),t) \\ \\ u_x(t)B_y(z(t),t)-u_y(t)B_x(z(t),t) \\ \\ \end{array} \right) \\ \\ &=& \boldsymbol{u}\times\boldsymbol{B}(z(t),t) \end{eqnarray} が得られる。

\(x\)成分だけ求めると \begin{eqnarray} p_x(t) &=& \frac{\partial \mathscr{L}(t)}{\partial \left(\frac{dz_x(t)}{dt}\right)} \\ \\ &=& \frac{\partial \mathscr{L}(t)}{\partial u_x(t)} \\ \\ &=& \frac{\partial }{\partial u_x(t)}\left(-mc^2\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}+e\boldsymbol{u}\cdot\boldsymbol{A}(\boldsymbol{z}(t),t)-eA_0(\boldsymbol{z}(t),t)+\mathscr{L}_F(t)\right) \\ \\ &=& \frac{\partial }{\partial u_x(t)}\left(-mc^2\sqrt{1-\frac{1}{c^2}(u_x^2+u_y^2+u_z^2)}+e(u_xA_x+u_yA_y+u_zA_z)-eA_0(\boldsymbol{z}(t),t)+\mathscr{L}_F(t)\right) \\ \\ &=& -mc^2\frac{-\frac{1}{c^2}u_x}{\sqrt{1-\frac{1}{c^2}(u_x^2+u_y^2+u_z^2)}}+eA_x\\ \\ &=& \frac{mu_x}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2}}+eA_x\\ \\ \end{eqnarray} と導出できる。

\begin{eqnarray} \pi_i(\boldsymbol{x}_{\alpha},t) &=& \frac{\partial \mathscr{L}(t)}{\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial t}\right)} \\ \\ &=& \frac{\partial \mathscr{L}(t)}{ic\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial ict}\right)} \\ \\ &=& \frac{\partial \mathscr{L}(t)}{ic\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial x_4}\right)} \\ \\ &=& \frac{\partial }{ic\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial x_4}\right)}\left(\mathscr{L}_M(t)+\mathscr{L}_I(t)+\mathscr{L}_F(t)\right) \\ \\ &=& \frac{\partial }{ic\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial x_4}\right)}\mathscr{L}_F(t)&...&\mathscr{L}_M(t),\mathscr{L}_I(t)\text{は}\frac{\partial A_i}{\partial x_4}\text{の項を持たないため} \\ \\ &=& \frac{\partial }{ic\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial x_4}\right)}\displaystyle\sum_{\beta}\left(-\frac{\varepsilon_0}{4}F_{\mu\nu}(\boldsymbol{x}_{\beta},t)F_{\mu\nu}(\boldsymbol{x}_{\beta},t)\right)\Delta^3x_{\beta} \\ \\ &=& -\frac{\varepsilon_0}{4ic}\frac{\partial }{\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial x_4}\right)}\displaystyle\sum_{\beta}\left(F_{\mu\nu}(\boldsymbol{x}_{\beta},t)F_{\mu\nu}(\boldsymbol{x}_{\beta},t)\right)\Delta^3x_{\beta} \\ \\ &=& -\frac{\varepsilon_0}{4ic}\frac{\partial }{\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial x_4}\right)}\displaystyle\sum_{\beta}\left(\frac{A_{\nu}(\boldsymbol{x}_{\beta},t)}{\partial x_{\mu}}-\frac{A_{\mu}(\boldsymbol{x}_{\beta},t)}{\partial x_{\nu}}\right)\left(\frac{A_{\nu}(\boldsymbol{x}_{\beta},t)}{\partial x_{\mu}}-\frac{A_{\mu}(\boldsymbol{x}_{\beta},t)}{\partial x_{\nu}}\right)\Delta^3x_{\beta} \\ \\ &=& -\frac{\varepsilon_0}{4ic}\frac{\partial }{\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial x_4}\right)}\displaystyle\sum_{\beta}\left(\frac{A_{\nu}(\boldsymbol{x}_{\beta},t)}{\partial x_{\mu}}-\frac{A_{\mu}(\boldsymbol{x}_{\beta},t)}{\partial x_{\nu}}\right)^2\Delta^3x_{\beta} \\ \\ &=& -\frac{\varepsilon_0}{4ic}\frac{\partial }{\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial x_4}\right)}\left(\frac{A_{\nu}(\boldsymbol{x}_{\alpha},t)}{\partial x_{\mu}}-\frac{A_{\mu}(\boldsymbol{x}_{\alpha},t)}{\partial x_{\nu}}\right)^2\Delta^3x_{\alpha}&...&\text{偏微分において少なくとも}x_{\alpha}\text{を含まない項は定数として扱えるため} \\ \\ &=& -\frac{\varepsilon_0}{4ic}2\left(\frac{A_{\nu}(\boldsymbol{x}_{\alpha},t)}{\partial x_{\mu}}-\frac{A_{\mu}(\boldsymbol{x}_{\alpha},t)}{\partial x_{\nu}}\right)\frac{\partial }{\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial x_4}\right)}\left(\frac{A_{\nu}(\boldsymbol{x}_{\alpha},t)}{\partial x_{\mu}}-\frac{A_{\mu}(\boldsymbol{x}_{\alpha},t)}{\partial x_{\nu}}\right)\Delta^3x_{\alpha}& \\ \\ &=& -\frac{\varepsilon_0}{2ic}F_{\mu\nu}(\boldsymbol{x}_{\alpha}(t),t)\frac{\partial }{\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial x_4}\right)}\left(\frac{A_{\nu}(\boldsymbol{x}_{\alpha},t)}{\partial x_{\mu}}-\frac{A_{\mu}(\boldsymbol{x}_{\alpha},t)}{\partial x_{\nu}}\right)\Delta^3x_{\alpha}& \\ \\ &=& -\frac{\varepsilon_0}{2ic}F_{\mu\nu}(\boldsymbol{x}_{\alpha}(t),t)\frac{\partial }{\partial \left(\frac{A_i(\boldsymbol{x}_{\alpha},t)}{\partial x_4}\right)}\left(\frac{cA_{\nu}(\boldsymbol{x}_{\alpha},t)}{\partial x_{\mu}}-\frac{cA_{\mu}(\boldsymbol{x}_{\alpha},t)}{\partial x_{\nu}}\right)\Delta^3x_{\alpha}&...&\text{ここで}i=x,y,z\text{であることから}\nu,\mu=1,2,3\text{になる必要がある。}\\&&&& \text{従ってp.385式(3.43)より}c\text{をかけ、}A\text{の添え字を}x,y,z\text{とした。i=0についてはp.424下に記載がある。} \\ \\ &=& -\frac{c\varepsilon_0}{2ic}F_{\mu\nu}(\boldsymbol{x}_{\alpha}(t),t)\left(\delta_{i\nu}\delta_{4\mu}-\delta_{i\mu}\delta_{4\nu}\right)\Delta^3x_{\alpha}&...&\text{括弧の中(微分する関数)は一項目が}\nu=i,\mu=4\text{、二項目が}\mu=i,\nu=4\text{の時のみ微分結果が}1\text{になる。} \\ \\ &=& -\frac{\varepsilon_0}{2i}F_{\mu\nu}(\boldsymbol{x}_{\alpha}(t),t)\left(\delta_{i\nu}\delta_{4\mu}-\delta_{i\mu}\delta_{4\nu}\right)\Delta^3x_{\alpha}&\\ \\ &=& -\frac{\varepsilon_0}{2i}\displaystyle\sum_{\nu\mu}^3F_{\mu\nu}(\boldsymbol{x}_{\alpha}(t),t)\left(\delta_{i\nu}\delta_{4\mu}-\delta_{i\mu}\delta_{4\nu}\right)\Delta^3x_{\alpha}&\\ \\ &=& -\frac{\varepsilon_0}{2i}\left(F_{i4}(\boldsymbol{x}_{\alpha}(t),t)-F_{4i}(\boldsymbol{x}_{\alpha}(t),t)\right)\Delta^3x_{\alpha}&\\ \\ &=& -\frac{\varepsilon_0}{2i}\left(F_{i4}(\boldsymbol{x}_{\alpha}(t),t)+F_{i4}(\boldsymbol{x}_{\alpha}(t),t)\right)\Delta^3x_{\alpha}&...&F_{\mu\nu}\text{は反対称であるため(p.383上)}\\ \\ &=& -\frac{\varepsilon_0}{i}F_{i4}(\boldsymbol{x}_{\alpha}(t),t)\Delta^3x_{\alpha}&\\ \\ &=& -\frac{\varepsilon_0}{i}(-i)E_{i}(\boldsymbol{x}_{\alpha}(t),t)\Delta^3x_{\alpha}&...&\text{p.381式(3.32)より}\\ \\ &=& -\varepsilon_0E_{i}(\boldsymbol{x}_{\alpha}(t),t)\Delta^3x_{\alpha}&\\ \\ \end{eqnarray} と導出できる。

二行目から三行目へは \begin{eqnarray} \left(\boldsymbol{p}(t)-e\boldsymbol{A}\right)\cdot\boldsymbol{u}(t) &=& \frac{m\boldsymbol{u}(t)}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}}\cdot\boldsymbol{u}(t) &...&\text{式(2.24)を利用} \\ \\ &=& \frac{m\boldsymbol{u}^2(t)}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)}} & \\ \\ \end{eqnarray} を用いて導出できる。
三行目から四行目へは \begin{eqnarray} \frac{m\boldsymbol{u}^2(t)}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)} }+mc^2\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)} &=& \frac{m\boldsymbol{u}^2(t)}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t)} }+\frac{mc^2(1-\frac{1}{c^2}\boldsymbol{u}^2(t))}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t) } } \\ \\ &=& \frac{mc^2}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2(t) } } \\ \\ \end{eqnarray} を用いて導出できる。

式(2.24)より \begin{eqnarray} &&\boldsymbol{p}&=&\frac{m\boldsymbol{u}}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2}}+e\boldsymbol{A} \\ \\ &\Rightarrow& (\boldsymbol{p}-e\boldsymbol{A})^2&=&\left(\frac{m\boldsymbol{u}}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2}}\right)^2 \\ \\ &&&=& \frac{m^2\boldsymbol{u}^2}{1-\frac{1}{c^2}\boldsymbol{u}^2}\\ \\ &&&=& \frac{m^2\boldsymbol{u}^2+m^2c^2\left(1-\frac{1}{c^2}\boldsymbol{u}^2\right)}{1-\frac{1}{c^2}\boldsymbol{u}^2}-m^2c^2\\ \\ &&&=& \frac{m^2c^2}{1-\frac{1}{c^2}\boldsymbol{u}^2}-m^2c^2\\ \\ &\Leftrightarrow& (\boldsymbol{p}-e\boldsymbol{A})^2+m^2c^2 &=& \frac{m^2c^2}{1-\frac{1}{c^2}\boldsymbol{u}^2} \end{eqnarray} と示される。

式(2.27)の \begin{eqnarray} \frac{mc^2}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2}} \end{eqnarray} に先ほど求めた \begin{eqnarray} (\boldsymbol{p}-e\boldsymbol{A})^2+m^2c^2 &=& \frac{m^2c^2}{1-\frac{1}{c^2}\boldsymbol{u}^2} \\ \\ \Leftrightarrow \frac{mc}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2}}&=&\pm \sqrt{(\boldsymbol{p}-e\boldsymbol{A})^2+m^2c^2} \\ \\ \Leftrightarrow \frac{mc^2}{\sqrt{1-\frac{1}{c^2}\boldsymbol{u}^2}}&=&\pm c\sqrt{(\boldsymbol{p}-e\boldsymbol{A})^2+m^2c^2} \\ \\ \end{eqnarray} を代入することで得られる。

\begin{eqnarray} H_{F} &=& \int_Vd^3x\left(\boldsymbol{\Pi}(\boldsymbol{x})\cdot\frac{\partial \boldsymbol{A}(\boldsymbol{x})}{\partial t}-L_F\right)&...&\text{式(2.29)} \\ \\ &=& \int_Vd^3x\left(-\varepsilon_0\boldsymbol{E}(\boldsymbol{x})\cdot\left(-\boldsymbol{E}(\boldsymbol{x})-\text{grad}A_0(\boldsymbol{x})\right)-\frac{1}{2}\left(\boldsymbol{D}(\boldsymbol{x})\cdot\boldsymbol{E}(\boldsymbol{x})-\boldsymbol{H}(\boldsymbol{x})\cdot\boldsymbol{B}(\boldsymbol{x})\right)\right)& \\ \\ &=& \int_Vd^3x\left(-\boldsymbol{D}(\boldsymbol{x})\cdot\left(-\boldsymbol{E}(\boldsymbol{x})-\text{grad}A_0(\boldsymbol{x})\right)-\frac{1}{2}\left(\boldsymbol{D}(\boldsymbol{x})\cdot\boldsymbol{E}(\boldsymbol{x})-\boldsymbol{H}(\boldsymbol{x})\cdot\boldsymbol{B}(\boldsymbol{x})\right)\right)& \\ \\ &=& \int_Vd^3x\left(\boldsymbol{D}(\boldsymbol{x})\cdot\left(-\text{grad}A_0(\boldsymbol{x})\right)+\frac{1}{2}\left(\boldsymbol{D}(\boldsymbol{x})\cdot\boldsymbol{E}(\boldsymbol{x})+\boldsymbol{H}(\boldsymbol{x})\cdot\boldsymbol{B}(\boldsymbol{x})\right)\right)& \\ \\ &=& \int_Vd^3x\boldsymbol{D}(\boldsymbol{x})\cdot\text{grad}A_0(\boldsymbol{x})+\frac{1}{2}\int_Vd^3x\left(\boldsymbol{D}(\boldsymbol{x})\cdot\boldsymbol{E}(\boldsymbol{x})+\boldsymbol{H}(\boldsymbol{x})\cdot\boldsymbol{B}(\boldsymbol{x})\right)& \\ \\ &=& -\int_Vd^3x\text{div}\boldsymbol{D}(\boldsymbol{x})A_0(\boldsymbol{x})+\frac{1}{2}\int_Vd^3x\left(\boldsymbol{D}(\boldsymbol{x})\cdot\boldsymbol{E}(\boldsymbol{x})+\boldsymbol{H}(\boldsymbol{x})\cdot\boldsymbol{B}(\boldsymbol{x})\right)&...&(1) \\ \\ &=& -\int_Vd^3xe\delta^3(\boldsymbol{x}-\boldsymbol{z}(t))A_0(\boldsymbol{x})+\frac{1}{2}\int_Vd^3x\left(\boldsymbol{D}(\boldsymbol{x})\cdot\boldsymbol{E}(\boldsymbol{x})+\boldsymbol{H}(\boldsymbol{x})\cdot\boldsymbol{B}(\boldsymbol{x})\right)&...&\text{p.423下より} \\ \\ &=& -eA_0(\boldsymbol{z}(t))+\frac{1}{2}\int_Vd^3x\left(\boldsymbol{D}(\boldsymbol{x})\cdot\boldsymbol{E}(\boldsymbol{x})+\boldsymbol{H}(\boldsymbol{x})\cdot\boldsymbol{B}(\boldsymbol{x})\right)&...&\text{デルタ関数の積分より} \\ \\ \end{eqnarray} として得られる。

1. (1)部分積分を利用
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p.445式(A・44)より \begin{eqnarray} \text{div}(\boldsymbol{D}A_0)&=&(\text{div}\boldsymbol{D})A_0+\boldsymbol{D}\cdot\text{grad}A_0 \end{eqnarray} と変形できる。両辺を積分すると \begin{eqnarray} &&\int_Vd^3x\text{div}(\boldsymbol{D}A_0)&=&\int_Vd^3x(\text{div}\boldsymbol{D})A_0+\int_Vd^3x\boldsymbol{D}\cdot\text{grad}A_0 \\ \\ &\Leftrightarrow& \int_SdS(\boldsymbol{D}A_0)\cdot \boldsymbol{n}&=&\int_Vd^3x(\text{div}\boldsymbol{D})A_0+\int_Vd^3x\boldsymbol{D}\cdot\text{grad}A_0&...&\text{p.445式(A・42)Gaussの定理より} \\ \\ &\Leftrightarrow& 0&=&\int_Vd^3x(\text{div}\boldsymbol{D})A_0+\int_Vd^3x\boldsymbol{D}\cdot\text{grad}A_0&...&\text{電子によって作られる電場と静電ポテンシャル(\(A_0\))なので、無限に大きな領域の表面では}0\text{になるとした。} \\ \\ &\Leftrightarrow& \int_Vd^3x\boldsymbol{D}(\boldsymbol{x})\cdot\text{grad}A_0(\boldsymbol{x})&=&-\int_Vd^3x(\text{div}\boldsymbol{D}(\boldsymbol{x}))A_0(\boldsymbol{x}) \\ \\ \end{eqnarray} が得られる。

はじめに\(\mu=1,2,3\)の時の\(x\)成分を考えると \begin{eqnarray} &&\Box A_{1}(\boldsymbol{x},t)-\frac{\partial}{\partial x}\left(\frac{A_{\nu}(\boldsymbol{x},t)}{x_{\nu}}\right)&=&-\frac{1}{c\varepsilon_0}j_x(\boldsymbol{x},t) \\ \\ &\Leftrightarrow& \Box cA_{x}(\boldsymbol{x},t)-\frac{\partial}{\partial x}\left(\frac{\partial A_{1}(\boldsymbol{x},t)}{\partial x_{1} }+\frac{\partial A_{2}(\boldsymbol{x},t)}{\partial x_{2} }+\frac{\partial A_{3}(\boldsymbol{x},t)}{\partial x_{3} }+\frac{\partial A_{4}(\boldsymbol{x},t)}{\partial x_{4} }\right)&=&-\frac{1}{c\varepsilon_0}i_x(\boldsymbol{x},t)&...&\text{p.378式(3.21)、p.385式(3.43)より} \\ \\ &\Leftrightarrow& \Box cA_{x}(\boldsymbol{x},t)-\frac{\partial}{\partial x}\left(\frac{c\partial A_x(\boldsymbol{x},t)}{\partial x }+\frac{c\partial A_y(\boldsymbol{x},t)}{\partial y }+\frac{\partial cA_z(\boldsymbol{x},t)}{\partial z }+\frac{\partial iA_{0}(\boldsymbol{x},t)}{\partial ix_{0} }\right)&=&-\frac{\mu_0}{c\mu_0\varepsilon_0}i_x(\boldsymbol{x},t)& \\ \\ &\Leftrightarrow& \Box cA_{x}(\boldsymbol{x},t)-\frac{\partial}{\partial x}\left(\frac{c\partial A_x(\boldsymbol{x},t)}{\partial x }+\frac{c\partial A_y(\boldsymbol{x},t)}{\partial y }+\frac{c\partial A_z(\boldsymbol{x},t)}{\partial z }+\frac{\partial A_{0}(\boldsymbol{x},t)}{\partial ct }\right)&=&-\frac{c^2\mu_0}{c}i_x(\boldsymbol{x},t)& \\ \\ &\Leftrightarrow& \Box cA_{x}(\boldsymbol{x},t)-c\frac{\partial}{\partial x}\left(\frac{\partial A_x(\boldsymbol{x},t)}{\partial x }+\frac{\partial A_y(\boldsymbol{x},t)}{\partial y }+\frac{\partial A_z(\boldsymbol{x},t)}{\partial z }+\frac{1}{c^2}\frac{\partial A_{0}(\boldsymbol{x},t)}{\partial t }\right)&=&-c\mu_0i_x(\boldsymbol{x},t)& \\ \\ &\Leftrightarrow& \Box A_{x}(\boldsymbol{x},t)-\frac{\partial}{\partial x}\left(\frac{\partial A_x(\boldsymbol{x},t)}{\partial x }+\frac{\partial A_y(\boldsymbol{x},t)}{\partial y }+\frac{\partial A_z(\boldsymbol{x},t)}{\partial z }+\frac{1}{c^2}\frac{\partial A_{0}(\boldsymbol{x},t)}{\partial t }\right)&=&-\mu_0i_x(\boldsymbol{x},t)& \\ \\ &\Leftrightarrow& \Box A_{x}(\boldsymbol{x},t)-\frac{\partial}{\partial x}\left(\text{div}\boldsymbol{A}+\frac{1}{c^2}\frac{\partial A_{0}(\boldsymbol{x},t)}{\partial t }\right)&=&-\mu_0i_x(\boldsymbol{x},t)& \\ \\ \end{eqnarray} より、三次元で考えると \begin{eqnarray} \Box \boldsymbol{A}(\boldsymbol{x},t)-\text{grad}\left(\text{div}\boldsymbol{A}+\frac{1}{c^2}\frac{\partial A_{0}(\boldsymbol{x},t)}{\partial t }\right)&=&-\mu_0\boldsymbol{i}(\boldsymbol{x},t)& \\ \\ \end{eqnarray} が得られる。
\(\mu=4\)を考えると \begin{eqnarray} &&\Box A_{4}(\boldsymbol{x},t)-\frac{\partial}{\partial x_4}\left(\frac{A_{\nu}(\boldsymbol{x},t)}{x_{\nu}}\right)&=&-\frac{1}{c\varepsilon_0}j_4(\boldsymbol{x},t) \\ \\ &\Leftrightarrow& \Box iA_{0}(\boldsymbol{x},t)-\frac{\partial}{\partial ict}\left(\frac{\partial A_{1}(\boldsymbol{x},t)}{\partial x_{1} }+\frac{\partial A_{2}(\boldsymbol{x},t)}{\partial x_{2} }+\frac{\partial A_{3}(\boldsymbol{x},t)}{\partial x_{3} }+\frac{\partial A_{4}(\boldsymbol{x},t)}{\partial x_{4} }\right)&=&-\frac{1}{c\varepsilon_0}ic\rho(\boldsymbol{x},t)&...&\text{p.378式(3.21)、p.385式(3.43)より} \\ \\ &\Leftrightarrow& \Box A_{0}(\boldsymbol{x},t)+\frac{\partial}{\partial ct}\left(\frac{c\partial A_x(\boldsymbol{x},t)}{\partial x }+\frac{c\partial A_y(\boldsymbol{x},t)}{\partial y }+\frac{\partial cA_z(\boldsymbol{x},t)}{\partial z }+\frac{\partial iA_{0}(\boldsymbol{x},t)}{\partial ix_{0} }\right)&=&-\frac{1}{\varepsilon_0}\rho(\boldsymbol{x},t)& \\ \\ &\Leftrightarrow& \Box A_{0}(\boldsymbol{x},t)+c\frac{\partial}{\partial ct}\left(\frac{\partial A_x(\boldsymbol{x},t)}{\partial x }+\frac{\partial A_y(\boldsymbol{x},t)}{\partial y }+\frac{\partial A_z(\boldsymbol{x},t)}{\partial z }+\frac{1}{c^2}\frac{\partial A_{0}(\boldsymbol{x},t)}{\partial t }\right)&=&-\frac{1}{\varepsilon_0}\rho(\boldsymbol{x},t)& \\ \\ &\Leftrightarrow& \Box A_{0}(\boldsymbol{x},t)+\frac{\partial}{\partial t}\left(\frac{\partial A_x(\boldsymbol{x},t)}{\partial x }+\frac{\partial A_y(\boldsymbol{x},t)}{\partial y }+\frac{\partial A_z(\boldsymbol{x},t)}{\partial z }+\frac{1}{c^2}\frac{\partial A_{0}(\boldsymbol{x},t)}{\partial t }\right)&=&-\frac{1}{\varepsilon_0}\rho(\boldsymbol{x},t)& \\ \\ &\Leftrightarrow& \Box A_{0}(\boldsymbol{x},t)+\frac{\partial}{\partial t}\left(\text{div}\boldsymbol{A}+\frac{1}{c^2}\frac{\partial A_{0}(\boldsymbol{x},t)}{\partial t }\right)&=&-\frac{1}{\varepsilon_0}\rho(\boldsymbol{x},t)& \\ \\ &\Leftrightarrow& \Box A_{0}(\boldsymbol{x},t)+\frac{\partial}{\partial t}\text{div}\boldsymbol{A}-\frac{\partial}{\partial t}\frac{1}{c^2}\frac{\partial A_{0}(\boldsymbol{x},t)}{\partial t }&=&-\frac{1}{\varepsilon_0}\rho(\boldsymbol{x},t)& \\ \\ &\Leftrightarrow& \left(\Delta-\frac{1}{c^2}\frac{\partial^2}{\partial t^2}\right) A_{0}(\boldsymbol{x},t)+\frac{\partial}{\partial t}\text{div}\boldsymbol{A}+\frac{1}{c^2}\frac{\partial^2 A_{0}(\boldsymbol{x},t)}{\partial t^2 }&=&-\frac{1}{\varepsilon_0}\rho(\boldsymbol{x},t)&...&\text{p.385式(3.44)より} \\ \\ &\Leftrightarrow& \Delta A_{0}(\boldsymbol{x},t)+\frac{\partial}{\partial t}\text{div}\boldsymbol{A}&=&-\frac{1}{\varepsilon_0}\rho(\boldsymbol{x},t)&\\ \\ \end{eqnarray} と導ける。

式(2.34)はp.47の式(3.9)と等しい。加えて、今回の式はp.48式(3.10)の\(\boldsymbol{A}',\phi'\)の導入と等しい\(u\to\lambda\)ため、式(2.34)は満たされている。

\begin{eqnarray} \boldsymbol{H}\cdot\boldsymbol{B} &=& \frac{1}{\mu_0}\mu_0\boldsymbol{H}\cdot\boldsymbol{B} \\ \\ &=& \frac{1}{\mu_0}\boldsymbol{B}\cdot\boldsymbol{B} \\ \\ &=& \frac{1}{\mu_0}(\boldsymbol{B})^2 \\ \\ &=& \frac{1}{\mu_0}(\text{rot}\boldsymbol{A}^c)^2 \\ \\ \end{eqnarray} の変形を用いている。

三行目から四行目の式変換は部分積分を用いている。 \begin{eqnarray} &&\text{div}([2\boldsymbol{E}_1+\boldsymbol{E}_2]A_0^c)=[2\boldsymbol{E}_1+\boldsymbol{E}_2]\cdot\text{grad}A_0^c+(\text{div}[2\boldsymbol{E}_1+\boldsymbol{E}_2])A_0^c&...&\text{p.445式(A・44)より}\\ \\ &\Leftrightarrow& \int_Vd^3x\text{div}([2\boldsymbol{E}_1+\boldsymbol{E}_2]A_0^c)=\int_Vd^3x[2\boldsymbol{E}_1+\boldsymbol{E}_2]\cdot\text{grad}A_0^c+\int_Vd^3x(\text{div}[2\boldsymbol{E}_1+\boldsymbol{E}_2])A_0^c&\\ \\ &\Leftrightarrow& \oint_SdS([2\boldsymbol{E}_1+\boldsymbol{E}_2]A_0^c)\cdot\boldsymbol{n}=\int_Vd^3x[2\boldsymbol{E}_1+\boldsymbol{E}_2]\cdot\text{grad}A_0^c+\int_Vd^3x(\text{div}[2\boldsymbol{E}_1+\boldsymbol{E}_2])A_0^c&...&\text{Gaussの法則(p.445)より}\\ \\ &\Leftrightarrow& 0=\int_Vd^3x[2\boldsymbol{E}_1+\boldsymbol{E}_2]\cdot\text{grad}A_0^c+\int_Vd^3x(\text{div}[2\boldsymbol{E}_1+\boldsymbol{E}_2])A_0^c&...&\text{点電荷が作る電場であり、無限に広い領域をとるとその表面では0になるため}\\ \\ &\Leftrightarrow& \int_Vd^3x[2\boldsymbol{E}_1+\boldsymbol{E}_2]\cdot\text{grad}A_0^c=-\int_Vd^3x(\text{div}[2\boldsymbol{E}_1+\boldsymbol{E}_2])A_0^c&\\ \\ \end{eqnarray}

デルタ関数の微分を利用し \begin{eqnarray} \frac{\varepsilon_0}{2}\int_V\frac{1}{\varepsilon_0}\rho(x)A_0^c(x) &=& \frac{1}{2}\int_V\displaystyle\sum_{i=1}^Ne_i\delta^3(\boldsymbol{x}-\boldsymbol{z}_i(t))A_0^c(x) \\ \\ &=& \frac{1}{2}\int_V\displaystyle\sum_{i=1}^Ne_iA_0^c(\boldsymbol{z}_i(t))&...&\text{デルタ関数の積分より、}\boldsymbol{x}=\boldsymbol{z}_i\text{の項が残る。} \\ \\ \end{eqnarray} の式変形を用いた。

p.254式(1.19)やp.87の式(2.7)にp.427下の条件を加えて積分するとp.274上段の\(\phi(\boldsymbol{x},t)\)と同じ結果が得られる。それをすべての電子Nについて総和をとると\(A_0^c(\boldsymbol{z},t)\)が得られる。

\begin{eqnarray} \text{div}(\boldsymbol{A}^c\times\text{rot}\boldsymbol{A}^c) &=& \text{rot}\boldsymbol{A}^c\cdot\text{rot}\boldsymbol{A}^c-\boldsymbol{A}^c\cdot\text{rot}\text{rot}\boldsymbol{A}^c&...&\text{p.449式((7)(b))より} \end{eqnarray} が得られる。両辺を積分すると \begin{eqnarray} \int_Vd^3x\text{div}(\boldsymbol{A}^c\times\text{rot}\boldsymbol{A}^c) &=& \int_Vd^3x\text{rot}\boldsymbol{A}^c\cdot\text{rot}\boldsymbol{A}^c-\int_Vd^3x\boldsymbol{A}^c\cdot\text{rot}\text{rot}\boldsymbol{A}^c&\\ \\ \int_SdS(\boldsymbol{A}^c\times\text{rot}\boldsymbol{A}^c)\cdot\boldsymbol{n} &=& \int_Vd^3x\text{rot}\boldsymbol{A}^c\cdot\text{rot}\boldsymbol{A}^c-\int_Vd^3x\boldsymbol{A}^c\cdot\text{rot}\text{rot}\boldsymbol{A}^c&...&\text{p.445式(A・42)Gaussの定理より}\\ \\ \end{eqnarray} が得られる。

式(2.42)より \begin{eqnarray} \text{div}\boldsymbol{E}_1&=&0\\ \\ \Leftrightarrow\frac{\partial}{\partial x}E_{1x}+\frac{\partial}{\partial y}E_{1y}+\frac{\partial}{\partial z}E_{1z}&=&0 \end{eqnarray} であることから、\(E_x,E_y\)を決めると自動的に\(E_z\)が決まることが推測される。従って、\(x,y,z\)のうち二つの成分は独立に決められ、一つはの他の二つに従って決定する。
これは\(\boldsymbol{A}^c\)も同様のことが言える。

\(\boldsymbol{e}_x,\boldsymbol{e}_y,\boldsymbol{e}_z\)をそれぞれ\(x,y,z\)方向の単位ベクトルとする。また、\(\boldsymbol{k}=(k_x,k_y,k_z)\)とした。 \begin{eqnarray} \text{div}\boldsymbol{A}^c/\sqrt{\frac{1}{\varepsilon_0(2\pi)^3}} &=& \text{div}\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right] \\ \\ &=& \frac{\partial}{\partial x}\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_x\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right] \\ &+&\frac{\partial}{\partial y}\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_y\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right] \\ &+&\frac{\partial}{\partial z}\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_z\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right]&...&\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\text{の\(x\)成分は} \\&&&&\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_x\text{となることを利用} \\ \\ &=& \displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_x\left[C^{(\lambda)}(\boldsymbol{k},t)ik_xe^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)(-ik_x)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right] \\ &+&\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_y\left[C^{(\lambda)}(\boldsymbol{k},t)ik_ye^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)(-ik_y)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right] \\ &+&\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_z\left[C^{(\lambda)}(\boldsymbol{k},t)ik_ze^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)(-ik_z)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right] &...&(1)\\ \\ &=& \displaystyle\sum_{\lambda=1,2}\int d^3k\left(\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_x\left[C^{(\lambda)}(\boldsymbol{k},t)ik_xe^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)(-ik_x)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right]\right. \\ &&+\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_y\left[C^{(\lambda)}(\boldsymbol{k},t)ik_ye^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)(-ik_y)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right] \\ &&+\left.\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_z\left[C^{(\lambda)}(\boldsymbol{k},t)ik_ze^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)(-ik_z)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right]\right) &\\ \\ &=& \displaystyle\sum_{\lambda=1,2}\int d^3kC^{(\lambda)}(\boldsymbol{k},t)\left(ik_x\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_x+ik_y\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_y+ik_z\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_z\right)e^{i\boldsymbol{k}\cdot\boldsymbol{x}} \\ &+&\displaystyle\sum_{\lambda=1,2}\int d^3kC^{(\lambda)*}(\boldsymbol{k},t)\left(-ik_x\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_x-ik_y\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_y-ik_z\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\cdot\boldsymbol{e}_z\right)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}} \\ \\ &=& \displaystyle\sum_{\lambda=1,2}\int d^3kiC^{(\lambda)}(\boldsymbol{k},t)(\boldsymbol{k}\cdot\boldsymbol{e}^{(\lambda)}(\boldsymbol{k}))e^{i\boldsymbol{k}\cdot\boldsymbol{x}} -\displaystyle\sum_{\lambda=1,2}\int d^3kiC^{(\lambda)*}(\boldsymbol{k},t)(\boldsymbol{k}\cdot\boldsymbol{e}^{(\lambda)}(\boldsymbol{k}))e^{-i\boldsymbol{k}\cdot\boldsymbol{x}} \\ \\ &=& \displaystyle\sum_{\lambda=1,2}\int d^3kiC^{(\lambda)}(\boldsymbol{k},t)\cdot 0\cdot e^{i\boldsymbol{k}\cdot\boldsymbol{x}} -\displaystyle\sum_{\lambda=1,2}\int d^3kiC^{(\lambda)*}(\boldsymbol{k},t)\cdot 0\cdot e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}&...&\text{図(2.1)より}\boldsymbol{k}\cdot\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})=0 \\ \\ &=& 0 \end{eqnarray} が得られることから、\(\text{div}\boldsymbol{A}^c=0\)が自動的に満たされている。
また、\(\text{div}\boldsymbol{E}_1\)については\(C^{(\lambda)}(\boldsymbol{k},t),C^{(\lambda)*}(\boldsymbol{k},t)\)の係数が、\(\text{div}\boldsymbol{A}^c\)の中の\(C^{(\lambda)}(\boldsymbol{k},t),C^{(\lambda)*}(\boldsymbol{k},t)\)の係数の定数倍であるため、同様の結果が得られる。

1. (1)\(\frac{\partial}{\partial x}e^{i\boldsymbol{k}\cdot\boldsymbol{x}},\frac{\partial}{\partial y}e^{i\boldsymbol{k}\cdot\boldsymbol{x}},\frac{\partial}{\partial z}e^{i\boldsymbol{k}\cdot\boldsymbol{x}}\)の計算
▼ 詳細を開く
\(\frac{\partial}{\partial x}e^{i\boldsymbol{k}\cdot\boldsymbol{x}}\)だけ計算すると \begin{eqnarray} \frac{\partial}{\partial x}e^{i\boldsymbol{k}\cdot\boldsymbol{x}} &=& \frac{\partial}{\partial x}e^{i(k_xx+k_yy+k_zz)} \\ \\ &=& ik_xe^{i(k_xx+k_yy+k_zz)} \\ \\ &=& ik_xe^{i\boldsymbol{k}\cdot\boldsymbol{x}} \\ \\ \end{eqnarray} と導出できる。他の計算も同様に計算できる。

\begin{eqnarray} \frac{\varepsilon_0}{2}\int_Vd^3x\boldsymbol{E}_1^2(\boldsymbol{x},t) &=& \frac{\varepsilon_0}{2}\int_Vd^3x\left(\sqrt{\frac{1}{\varepsilon_0(2\pi)^3}}\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})i\omega(\boldsymbol{k})\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}-C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right] \right)^2 \\ \\ &=& \frac{\varepsilon_0}{2}\frac{1}{\varepsilon_0(2\pi)^3}\int_Vd^3x\left(\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})i\omega(\boldsymbol{k})\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}-C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right] \right)^2 \\ \\ &=& \frac{\varepsilon_0}{2}\frac{1}{\varepsilon_0(2\pi)^3}\int_Vd^3x\left(\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})i\omega(\boldsymbol{k})\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}-C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right] \right) \\ &&\times\left(\displaystyle\sum_{\lambda^{\prime}=1,2}\int d^3k^{\prime}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})i\omega(\boldsymbol{k}^{\prime})\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}-C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right] \right) \\ \\ &=& \frac{1}{2(2\pi)^3}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\int_Vd^3x\left(\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})i\omega(\boldsymbol{k})\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}-C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right] \right) \\ &&\times\left(\int d^3k^{\prime}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})i\omega(\boldsymbol{k}^{\prime})\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}-C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right] \right) \\ \\ &=& \frac{1}{2(2\pi)^3}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\int_Vd^3x\left(\int\int d^3k d^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})i\omega(\boldsymbol{k})i\omega(\boldsymbol{k}^{\prime}) \right. \\ &&\left.\times\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}-C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right]\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}-C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right]\right) \\ \\ &=& \frac{1}{2(2\pi)^3}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\int_Vd^3x\left(\int\int d^3k d^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})i^2\omega(\boldsymbol{k})\omega(\boldsymbol{k}^{\prime}) \right. \\ &&\left.\times\left[C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i(\boldsymbol{k}+\boldsymbol{k}^{\prime})\cdot\boldsymbol{x}}-C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{i(\boldsymbol{k}-\boldsymbol{k}^{\prime})\cdot\boldsymbol{x}}-C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i(-\boldsymbol{k}+\boldsymbol{k}^{\prime})\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{-i(\boldsymbol{k}+\boldsymbol{k}^{\prime})\cdot\boldsymbol{x}}\right]\right) \\ \\ &=& \frac{-1}{2(2\pi)^3}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\left(\int\int d^3k d^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\omega(\boldsymbol{k})\omega(\boldsymbol{k}^{\prime}) \right. \\ &&\left.\times\left[C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)(2\pi)^3\delta(\boldsymbol{k}+\boldsymbol{k}^{\prime})-C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)(2\pi)^3\delta(\boldsymbol{k}-\boldsymbol{k}^{\prime})-C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)(2\pi)^3\delta(-\boldsymbol{k}+\boldsymbol{k}^{\prime})+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)(2\pi)^3\delta(-\boldsymbol{k}-\boldsymbol{k}^{\prime})\right]\right)...(1) \\ \\ &=& \frac{-(2\pi)^3}{2(2\pi)^3}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\left(\int\int d^3k d^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\omega(\boldsymbol{k})\omega(\boldsymbol{k}^{\prime}) \right. \\ &&\left.\times\left[C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\delta(\boldsymbol{k}+\boldsymbol{k}^{\prime})-C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)\delta(\boldsymbol{k}-\boldsymbol{k}^{\prime})-C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\delta(-\boldsymbol{k}+\boldsymbol{k}^{\prime})+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\delta(-\boldsymbol{k}-\boldsymbol{k}^{\prime})\right]\right) \\ \\ &=& \frac{-1}{2}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\left(\int\int d^3k d^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\omega(\boldsymbol{k})\omega(\boldsymbol{k}^{\prime}) \right. \\ &&\left.\times\left[C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\delta(\boldsymbol{k}+\boldsymbol{k}^{\prime})-C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)\delta(\boldsymbol{k}-\boldsymbol{k}^{\prime})-C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\delta(\boldsymbol{k}-\boldsymbol{k}^{\prime})+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\delta(\boldsymbol{k}+\boldsymbol{k}^{\prime})\right]\right)...\text{デルタ関数は偶関数であるから} \\ \\ &=& \frac{-1}{2}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\left(\int\int d^3k d^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\omega(\boldsymbol{k})\omega(\boldsymbol{k}^{\prime}) \right. \\ &&\left.\times\left[C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\delta(\boldsymbol{k}+\boldsymbol{k}^{\prime})-C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)\delta(\boldsymbol{k}-\boldsymbol{k}^{\prime})-C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\delta(\boldsymbol{k}-\boldsymbol{k}^{\prime})+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\delta(\boldsymbol{k}+\boldsymbol{k}^{\prime})\right]\right) \\ \\ &=& \frac{-1}{2}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\left(\int\int d^3k d^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\omega(\boldsymbol{k})\omega(\boldsymbol{k}^{\prime}) \right. \\ &&\left.\times\left[-\left\{C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\right\}\delta(\boldsymbol{k}-\boldsymbol{k}^{\prime})+\left\{C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\right\}\delta(\boldsymbol{k}+\boldsymbol{k}^{\prime})\right]\right) \\ \\ &=& \frac{-1}{2}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\left(-\int\int d^3k d^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\omega(\boldsymbol{k})\omega(\boldsymbol{k}^{\prime})\left\{C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\right\}\delta(\boldsymbol{k}-\boldsymbol{k}^{\prime}) \right. \\ &&\left.+\int\int d^3k d^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\omega(\boldsymbol{k})\omega(\boldsymbol{k}^{\prime})\left\{C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\right\}\delta(\boldsymbol{k}+\boldsymbol{k}^{\prime})\right) \\ \\ &=& \frac{-1}{2}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\left(-\int d^3k \boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k})\omega(\boldsymbol{k})\omega(\boldsymbol{k})\left\{C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k},t)\right\} \right. ...\text{デルタ関数の積分より}\boldsymbol{k}^{\prime}=\boldsymbol{k}\text{の項が残る}\\ &&\left.+\int d^3k \boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(-\boldsymbol{k})\omega(\boldsymbol{k})\omega(-\boldsymbol{k})\left\{C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(-\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(-\boldsymbol{k},t)\right\}\right) ...\text{デルタ関数の積分より}\boldsymbol{k}^{\prime}=-\boldsymbol{k}\text{の項が残る}\\ \\ &=& \frac{1}{2}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\left(\int d^3k \boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k})\omega(\boldsymbol{k})\omega(\boldsymbol{k})\left\{C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k},t)\right\} \right. \\ &&\left.-\int d^3k \boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(-\boldsymbol{k})\omega(\boldsymbol{k})\omega(-\boldsymbol{k})\left\{C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(-\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(-\boldsymbol{k},t)\right\}\right) \\ \\ &=& \frac{1}{2}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\int d^3k\left[ \boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k})\omega(\boldsymbol{k})^2\left\{C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k},t)\right\} \right. \\ &&\left.-\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(-\boldsymbol{k})\omega(\boldsymbol{k})\omega(\boldsymbol{k})\left\{C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(-\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(-\boldsymbol{k},t)\right\}\right]...\text{p.429下段より\(\omega\)は\(\boldsymbol{k}\)の大きさの関数なので}\omega(\boldsymbol{k})=\omega(-\boldsymbol{k}) \\ \\ &=& \frac{1}{2}\displaystyle\sum_{\lambda,\lambda^{\prime}=1,2}\int d^3k\left[ \boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k})\omega(\boldsymbol{k})^2\left\{C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k},t)\right\} \right. \\ &&\left.-\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(-\boldsymbol{k})\omega(\boldsymbol{k})^2\left\{C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(-\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(-\boldsymbol{k},t)\right\}\right]& \\ \\ \end{eqnarray} と式変形できる。

1. (1)デルタ関数の公式
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p.455(B・25)を三次元に拡張することで、 \begin{eqnarray} \int d^3ke^{i\boldsymbol{k}\cdot(\boldsymbol{x}-\boldsymbol{x}^{\prime})} &=& \int d^3ke^{ik_x(x-x^{\prime})+ik_y(y-y^{\prime})+ik_z(z-z^{\prime})} \\ \\ &=& \left(\int dke^{ik_x(x-x^{\prime})}\right)^3 \\ \\ &=& (2\pi)^3\delta(x-x^{\prime})\delta(y-y^{\prime})\delta(z-z^{\prime}) \\ \\ &=& (2\pi)^3\delta^3(\boldsymbol{x}-\boldsymbol{x}^{\prime}) \\ \\ \end{eqnarray} が得られる。

\begin{eqnarray} -\frac{\varepsilon_0c^2}{2}\int_Vd^3x\boldsymbol{A}^c(\boldsymbol{x},t)\Delta\boldsymbol{A}^c(\boldsymbol{x},t) &=& -\frac{\varepsilon_0c^2}{2}\int_Vd^3x\left(\sqrt{\frac{1}{\varepsilon_0(2\pi)^3}}\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right]\right. \\ &&\left.\Delta\left(\sqrt{\frac{1}{\varepsilon_0(2\pi)^3}}\displaystyle\sum_{\lambda^{\prime}=1,2}\int d^3k^{\prime}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}+C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right]\right) \right) \\ \\ &=& -\frac{\varepsilon_0c^2}{2}\int_Vd^3x\left(\sqrt{\frac{1}{\varepsilon_0(2\pi)^3}}\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right]\right. \\ &&\left.\left(\sqrt{\frac{1}{\varepsilon_0(2\pi)^3}}\displaystyle\sum_{\lambda^{\prime}=1,2}\int d^3k^{\prime}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})(-\boldsymbol{k}^{\prime 2})\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}+C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right]\right) \right)...(1) \\ \\ &=& -\frac{\varepsilon_0c^2}{2}\frac{1}{\varepsilon_0(2\pi)^3}\int_Vd^3x\left(\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right]\right. \\ &&\left.\left(\displaystyle\sum_{\lambda^{\prime}=1,2}\int d^3k^{\prime}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})(-\boldsymbol{k}^{\prime 2})\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}+C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right]\right) \right) \\ \\ &=& -\frac{c^2}{2(2\pi)^3}\int_Vd^3x\left(\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right]\right. \\ &&\left.\left(\displaystyle\sum_{\lambda^{\prime}=1,2}\int d^3k^{\prime}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})(-\boldsymbol{k}^{\prime 2})\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}+C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right]\right) \right) \\ \\ &=& \frac{c^2}{2(2\pi)^3}\int_Vd^3x\left(\displaystyle\sum_{\lambda=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right]\right. \\ &&\left.\left(\displaystyle\sum_{\lambda^{\prime}=1,2}\int d^3k^{\prime}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\boldsymbol{k}^{\prime 2}\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}+C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right]\right) \right) \\ \\ &=& \frac{c^2}{2(2\pi)^3}\int_Vd^3x\left(\displaystyle\sum_{\lambda\lambda^{\prime}=1,2}\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right]\right. \\ &&\left.\times\int d^3k^{\prime}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\boldsymbol{k}^{\prime 2}\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}+C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right] \right) \\ \\ &=& \frac{c^2}{2(2\pi)^3}\int_Vd^3x\left(\displaystyle\sum_{\lambda\lambda^{\prime}=1,2}\int\int d^3kd^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{k}^{\prime 2}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\right. \\ &&\left.\times\left[C^{(\lambda)}(\boldsymbol{k},t)e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right]\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}+C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right] \right) \\ \\ &=& \frac{c^2}{2(2\pi)^3}\int_Vd^3x\left(\displaystyle\sum_{\lambda\lambda^{\prime}=1,2}\int\int d^3kd^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{k}^{\prime 2}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\right. \\ &&\left.\times\left[C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i(\boldsymbol{k}+\boldsymbol{k}^{\prime})\cdot\boldsymbol{x}}+C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{i(\boldsymbol{k}-\boldsymbol{k}^{\prime})\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i(-\boldsymbol{k}+\boldsymbol{k}^{\prime})\cdot\boldsymbol{x}}+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{i(-\boldsymbol{k}-\boldsymbol{k}^{\prime})\cdot\boldsymbol{x}}\right] \right) \\ \\ &=& \frac{c^2}{2(2\pi)^3}\left(\displaystyle\sum_{\lambda\lambda^{\prime}=1,2}\int\int d^3kd^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{k}^{\prime 2}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\right. \\ &&\left.\times\left[C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)(2\pi)^3\delta^3(\boldsymbol{k}+\boldsymbol{k}^{\prime})+C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)(2\pi)^3\delta^3(\boldsymbol{k}-\boldsymbol{k}^{\prime})+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)(2\pi)^3\delta^3(-\boldsymbol{k}+\boldsymbol{k}^{\prime})+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)(2\pi)^3\delta^3(-\boldsymbol{k}-\boldsymbol{k}^{\prime})\right] \right)...(2) \\ \\ &=& \frac{c^2(2\pi)^3}{2(2\pi)^3}\left(\displaystyle\sum_{\lambda\lambda^{\prime}=1,2}\int\int d^3kd^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{k}^{\prime 2}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\right. \\ &&\left.\times\left[C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\delta^3(\boldsymbol{k}+\boldsymbol{k}^{\prime})+C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)\delta^3(\boldsymbol{k}-\boldsymbol{k}^{\prime})+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\delta^3(\boldsymbol{k}-\boldsymbol{k}^{\prime})+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)\delta^3(\boldsymbol{k}+\boldsymbol{k}^{\prime})\right] \right)...\text{デルタ関数は偶関数であるため} \\ \\ &=& \frac{c^2(2\pi)^3}{2(2\pi)^3}\left(\displaystyle\sum_{\lambda\lambda^{\prime}=1,2}\int\int d^3kd^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{k}^{\prime 2}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\right. \\ &&\left.\times\left[\left(C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)\right)\delta^3(\boldsymbol{k}+\boldsymbol{k}^{\prime})+\left(C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\right)\delta^3(\boldsymbol{k}-\boldsymbol{k}^{\prime})\right] \right) \\ \\ &=& \frac{c^2}{2}\displaystyle\sum_{\lambda\lambda^{\prime}=1,2}\left(\int\int d^3kd^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{k}^{\prime 2}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\left(C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)\right)\delta^3(\boldsymbol{k}+\boldsymbol{k}^{\prime})\right. \\ &&\left.+\int\int d^3kd^3k^{\prime}\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{k}^{\prime 2}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\left(C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)\right)\delta^3(\boldsymbol{k}-\boldsymbol{k}^{\prime}) \right) \\ \\ &=& \frac{c^2}{2}\displaystyle\sum_{\lambda\lambda^{\prime}=1,2}\left(\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})(-\boldsymbol{k})^{2}\boldsymbol{e}^{(\lambda^{\prime})}(-\boldsymbol{k})\left(C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(-\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(-\boldsymbol{k},t)\right)\right. ...\text{デルタ関数の積分より}\boldsymbol{k}^{\prime}=-\boldsymbol{k} \\ &&\left.+\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{k}^{ 2}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k})\left(C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k},t)\right) \right) ...\text{デルタ関数の積分より}\boldsymbol{k}^{\prime}=\boldsymbol{k}\\ \\ &=& \frac{c^2}{2}\displaystyle\sum_{\lambda\lambda^{\prime}=1,2}\left(\int d^3k\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{k}^{2}\boldsymbol{e}^{(\lambda^{\prime})}(-\boldsymbol{k})\left(C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(-\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(-\boldsymbol{k},t)\right)\right. \\ &&\left.+\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{k}^{ 2}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k})\left(C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k},t)\right) \right) \\ \\ &=& \frac{1}{2}\displaystyle\sum_{\lambda\lambda^{\prime}=1,2}\int d^3k(c\boldsymbol{k})^{2}\left\{\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(-\boldsymbol{k})\left(C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(-\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(-\boldsymbol{k},t)\right)\right. \\ &&\left.+\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k})\left(C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k},t)\right) \right\} \\ \\ &=& \frac{1}{2}\displaystyle\sum_{\lambda\lambda^{\prime}=1,2}\int d^3k(c\boldsymbol{k})^{2}\left\{\boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k})\left(C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(\boldsymbol{k},t)\right) \right. \\ &&\left.+ \boldsymbol{e}^{(\lambda)}(\boldsymbol{k})\boldsymbol{e}^{(\lambda^{\prime})}(-\boldsymbol{k})\left(C^{(\lambda)}(\boldsymbol{k},t)C^{(\lambda^{\prime})}(-\boldsymbol{k},t)+C^{(\lambda)*}(\boldsymbol{k},t)C^{(\lambda^{\prime})*}(-\boldsymbol{k},t)\right)\right\} \\ \\ \end{eqnarray} と式変形できる。

1. (2)\(\Delta\boldsymbol{A}^c(\boldsymbol{x},t)\)の計算
▼ 詳細を開く
微分に関係するのは\(e^{i\boldsymbol{k}\cdot\boldsymbol{x}},e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\)であるから、この二つに着目する。 \begin{eqnarray} \Delta e^{i\boldsymbol{k}\cdot\boldsymbol{x} } &=& \left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)e^{i\boldsymbol{k}\cdot\boldsymbol{x} } \\ \\ &=& \left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)e^{i(k_xx+k_yy+k_zz)} \\ \\ &=& \left( (ik_x)^2+(ik_y)^2+(ik_z)^2\right)e^{i(k_xx+k_yy+k_zz)} \\ \\ &=& -\left(k_x^2+k_y^2+k_z^2\right)e^{i\boldsymbol{k}\cdot\boldsymbol{x}} \\ \\ &=& -\boldsymbol{k}^2e^{i\boldsymbol{k}\cdot\boldsymbol{x} } \\ \\ \\ \Delta e^{-i\boldsymbol{k}\cdot\boldsymbol{x}} &=& \left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}} \\ \\ &=& \left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\right)e^{-i(k_xx+k_yy+k_zz)} \\ \\ &=& \left((-ik_x)^2+(-ik_y)^2+(-ik_z)^2\right)e^{-i(k_xx+k_yy+k_zz)} \\ \\ &=& -\left(k_x^2+k_y^2+k_z^2\right)e^{-i\boldsymbol{k}\cdot\boldsymbol{x}} \\ \\ &=& -\boldsymbol{k}^2e^{-i\boldsymbol{k}\cdot\boldsymbol{x}} \\ \\ \end{eqnarray} が得られる。これらを用いて、 \begin{eqnarray} \Delta\left(\sqrt{\frac{1}{\varepsilon_0(2\pi)^3}}\displaystyle\sum_{\lambda^{\prime}=1,2}\int d^3k^{\prime}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}+C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right]\right) &=& \sqrt{\frac{1}{\varepsilon_0(2\pi)^3}}\displaystyle\sum_{\lambda^{\prime}=1,2}\int d^3k^{\prime}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)(-\boldsymbol{k}^2)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}+C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)(-\boldsymbol{k}^2)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right] \\ \\ &=& \sqrt{\frac{1}{\varepsilon_0(2\pi)^3}}\displaystyle\sum_{\lambda^{\prime}=1,2}\int d^3k^{\prime}\boldsymbol{e}^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime})(-\boldsymbol{k}^2)\left[C^{(\lambda^{\prime})}(\boldsymbol{k}^{\prime},t)e^{i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}+C^{(\lambda^{\prime})*}(\boldsymbol{k}^{\prime},t)e^{-i\boldsymbol{k}^{\prime}\cdot\boldsymbol{x}}\right] \\ \\ \end{eqnarray} が得られる。


2. (2)デルタ関数の公式
▼ 詳細を開く
p.455(B・25)を三次元に拡張することで、 \begin{eqnarray} \int d^3ke^{i\boldsymbol{k}\cdot(\boldsymbol{x}-\boldsymbol{x}^{\prime})} &=& \int d^3ke^{ik_x(x-x^{\prime})+ik_y(y-y^{\prime})+ik_z(z-z^{\prime})} \\ \\ &=& \left(\int dke^{ik_x(x-x^{\prime})}\right)^3 \\ \\ &=& (2\pi)^3\delta(x-x^{\prime})\delta(y-y^{\prime})\delta(z-z^{\prime}) \\ \\ &=& (2\pi)^3\delta^3(\boldsymbol{x}-\boldsymbol{x}^{\prime}) \\ \\ \end{eqnarray} が得られる。

こちらの解説などを参考

\begin{eqnarray} T^{(c)}_{44} &=& \frac{\partial L_F}{\partial\left(\frac{\partial A_{\rho}}{\partial x_4} \right) }\frac{\partial A_{\rho}}{\partial x_{4}}-\delta_{44}F_{L} \\ \\ &=& \frac{\partial L_F}{\partial\left(\frac{\partial A_{\rho}}{\partial ict} \right) }\frac{\partial A_{\rho}}{\partial ict} -F_{L}\\ \\ &=& \frac{\partial L_F}{\partial\left(\frac{\partial A_{\rho}}{\partial t} \right) }\frac{\partial A_{\rho}}{\partial t} -F_{L}\\ \\ &=& \frac{\partial L_F}{\partial\left(\frac{\partial A_{1}}{\partial t} \right) }\frac{\partial A_{1}}{\partial t}+\frac{\partial L_F}{\partial\left(\frac{\partial A_{2}}{\partial t} \right) }\frac{\partial A_{2}}{\partial t}+\frac{\partial L_F}{\partial\left(\frac{\partial A_{3}}{\partial t} \right) }\frac{\partial A_{3}}{\partial t}+\frac{\partial L_F}{\partial\left(\frac{\partial A_{4}}{\partial t} \right) }\frac{\partial A_{4}}{\partial t} -F_{L}\\ \\ &=& \frac{\partial L_F}{\partial\left(\frac{\partial \boldsymbol{A}}{\partial t} \right) }\cdot\frac{\partial \boldsymbol{A}}{\partial t}+\frac{\partial L_F}{\partial\left(\frac{\partial A_{4}}{\partial t} \right) }\frac{\partial A_{4}}{\partial t} -F_{L}\\ \\ &=& \frac{\partial L_F}{\partial\left(\frac{\partial \boldsymbol{A}}{\partial t} \right) }\cdot\frac{\partial \boldsymbol{A}}{\partial t}+\frac{\partial L_F}{\partial\left(\frac{\partial A_{4}}{\partial ict} \right) }\frac{\partial A_{4}}{\partial ict} -F_{L}\\ \\ &=& \frac{\partial L_F}{\partial\left(\frac{\partial \boldsymbol{A}}{\partial t} \right) }\cdot\frac{\partial \boldsymbol{A}}{\partial t}+\frac{\partial L_F}{\partial\left(\frac{\partial A_{4}}{\partial x_4} \right) }\frac{\partial A_{4}}{\partial x_4} -F_{L}\\ \\ &=& \frac{\partial L_F}{\partial\left(\frac{\partial \boldsymbol{A}}{\partial t} \right) }\cdot\frac{\partial \boldsymbol{A}}{\partial t} -F_{L}&...&(1)\\ \\ \end{eqnarray} と得られる。

1. (1)\(\frac{\partial L_F}{\partial\left(\frac{\partial A_{4}}{\partial x_4} \right) }\frac{\partial A_{4}}{\partial x_4}=0\)の計算
▼ 詳細を開く
\begin{eqnarray} L_F &=& -\frac{\varepsilon_0}{4}F_{\mu\nu}F_{\mu\nu}&...&\text{式(2.2)より}\\ \\ &=& -\frac{\varepsilon_0}{4}(\frac{\partial A_{\nu}}{\partial x_{\mu}}-\frac{\partial A_{\mu}}{\partial x_{\nu}})(\frac{\partial A_{\nu}}{\partial x_{\mu}}-\frac{\partial A_{\mu}}{\partial x_{\nu}})&...&\text{p.385式(3.42)より}\\ \\ &\Rightarrow& -\frac{\varepsilon_0}{4}(\frac{\partial A_{4}}{\partial x_{4}}-\frac{\partial A_{4}}{\partial x_{4}})(\frac{\partial A_{4}}{\partial x_{4}}-\frac{\partial A_{4}}{\partial x_{4}})&...&\frac{\partial A_{4}}{\partial x_{4}}\text{の係数を求めるために}\mu=4,\nu=4\text{を代入}\\ \\ &=& 0 \end{eqnarray} であることから、\(\frac{\partial A_{4}}{\partial x_{4}}\)の係数は0になることがわかる。そのため、\(L_F\)を\(\frac{\partial A_{4}}{\partial x_{4}}\)で微分しても\(0\)になる。

\begin{eqnarray} L_F &=& -\frac{\varepsilon_0}{4}F_{\mu\nu}F_{\mu\nu}&...&\text{式(2.2)より}\\ \\ &=& -\frac{\varepsilon_0}{4}(\frac{\partial A_{\nu}}{\partial x_{\mu}}-\frac{\partial A_{\mu}}{\partial x_{\nu}})(\frac{\partial A_{\nu}}{\partial x_{\mu}}-\frac{\partial A_{\mu}}{\partial x_{\nu}})&...&\text{p.385式(3.42)より}\\ \\ &\Rightarrow& -\frac{\varepsilon_0}{4}(\frac{\partial A_{4}}{\partial x_{4}}-\frac{\partial A_{4}}{\partial x_{4}})(\frac{\partial A_{4}}{\partial x_{4}}-\frac{\partial A_{4}}{\partial x_{4}})&...&\frac{\partial A_{4}}{\partial x_{4}}\text{の係数を求めるために}\mu=4,\nu=4\text{を代入}\\ \\ &=& 0 \end{eqnarray} であることから、\(\frac{\partial A_{4}}{\partial x_{4}}\)の係数は0になることがわかる。そのため、\(L_F\)を\(\frac{\partial A_{4}}{\partial x_{4}}\)で微分しても\(0\)になる。

p.423下段、式(2.2)、p.384式(3.39)、p.387式(3.52)より \begin{eqnarray} L_F &=& -\frac{\varepsilon_0}{4}F_{\mu\nu}F_{\mu\nu} \\ \\ &=& \frac{1}{2}(\boldsymbol{D}\cdot\boldsymbol{E}-\boldsymbol{H}\cdot\boldsymbol{B}) \end{eqnarray} であり、 \begin{eqnarray} L_F &=& \frac{1}{2}(\boldsymbol{D}\cdot\boldsymbol{E}-\boldsymbol{H}\cdot\boldsymbol{B}) \\ \\ &=& \frac{1}{2}(\varepsilon_0\boldsymbol{E}\cdot\boldsymbol{E}-\boldsymbol{H}\cdot\boldsymbol{B}) \\ \\ &=& \frac{1}{2}(\varepsilon_0\boldsymbol{E}^2-\boldsymbol{H}\cdot\boldsymbol{B}) \\ \\ &=& \frac{1}{2}(\varepsilon_0(-\text{grad}\phi-\frac{\partial \boldsymbol{A}}{\partial t})^2-\boldsymbol{H}\cdot\boldsymbol{B}) \\ \\ \end{eqnarray} と書けることから、 \begin{eqnarray} \frac{\partial L_F}{\partial\left(\frac{\partial \boldsymbol{A} }{\partial x_4}\right)} &=& \frac{\partial L_F}{\partial\left(\frac{\partial \boldsymbol{A} }{\partial ict}\right)} \\ \\ &=& ic\frac{\partial }{\partial\left(\frac{\partial \boldsymbol{A} }{\partial t}\right)}L_F \\ \\ &=& ic\frac{\partial }{\partial\left(\frac{\partial \boldsymbol{A} }{\partial t}\right)}\frac{1}{2}(\varepsilon_0(-\text{grad}\phi-\frac{\partial \boldsymbol{A}}{\partial t})^2-\boldsymbol{H}\cdot\boldsymbol{B}) \\ \\ &=& ic\frac{\partial }{\partial\left(\frac{\partial \boldsymbol{A} }{\partial t}\right)}\frac{1}{2}\varepsilon_0(-\text{grad}\phi-\frac{\partial \boldsymbol{A}}{\partial t})^2 &...&\frac{\partial \boldsymbol{A} }{\partial t}\text{を含まない項を\(0\)にした。}\\ \\ &=& ic\frac{1}{2}\varepsilon_0\cdot (-2)(-\text{grad}\phi-\frac{\partial \boldsymbol{A}}{\partial t}) \\ \\ &=& -ic\varepsilon_0(-\text{grad}\phi-\frac{\partial \boldsymbol{A}}{\partial t}) \\ \\ &=& -ic\varepsilon_0\boldsymbol{E} \\ \\ &=& ic\boldsymbol{\Pi} \\ \\ \end{eqnarray} p.423中段の\(\boldsymbol{\Pi}(\boldsymbol{x})=-\varepsilon_0\boldsymbol{E}(\boldsymbol{x})\)を用いた。

\begin{eqnarray} &&\text{div}(A_i\varepsilon_0\boldsymbol{E})=A_i(\text{div}\varepsilon_0\boldsymbol{E})+\text{grad}A_i\cdot(\varepsilon_0\boldsymbol{E}) &...&\text{p.445(A・44)} \\ \\ &\Leftrightarrow& \int_Vd^3x\text{div}(A_i\varepsilon_0\boldsymbol{E})=\int_Vd^3xA_i(\text{div}\varepsilon_0\boldsymbol{E})+\int_Vd^3x\text{grad}A_i\cdot(\varepsilon_0\boldsymbol{E}) &...&\text{両辺を積分した} \\ \\ &\Leftrightarrow& \int_SdS(A_i\varepsilon_0\boldsymbol{E})\boldsymbol{n}=\int_Vd^3xA_i(\text{div}\varepsilon_0\boldsymbol{E})+\int_Vd^3x\text{grad}A_i\cdot(\varepsilon_0\boldsymbol{E}) &...&\text{p.445式(A・42)Gaussの定理より} \\ \\ &\Leftrightarrow& 0=\int_Vd^3xA_i(\text{div}\varepsilon_0\boldsymbol{E})+\int_Vd^3x\text{grad}A_i\cdot(\varepsilon_0\boldsymbol{E}) &...&\text{p.433上部で表面積分は消えると仮定している。} \\ \\ &\Leftrightarrow& \int_Vd^3x\text{grad}A_i\cdot(\varepsilon_0\boldsymbol{E})=-\int_Vd^3xA_i(\text{div}\varepsilon_0\boldsymbol{E}) & \\ \\ \end{eqnarray} であるから \begin{eqnarray} P^f_i &=& \int_Vd^3x\left[\nabla A_i\cdot\varepsilon_0\boldsymbol{E}+\frac{1}{c^2}(\boldsymbol{E}\times\boldsymbol{H})_i\right] \\ \\ &=& \int_Vd^3x\left[-A_i\text{div}\varepsilon_0\boldsymbol{E}+\frac{1}{c^2}(\boldsymbol{E}\times\boldsymbol{H})_i\right] \\ \\ &=& \int_Vd^3x\left[-A_ie\delta^3(\boldsymbol{x}-\boldsymbol{z}(t))+\frac{1}{c^2}(\boldsymbol{E}\times\boldsymbol{H})_i\right] &...&\text{p.433上部より}\\ \\ &=& \int_Vd^3x\left[-A_ie\delta^3(\boldsymbol{x}-\boldsymbol{z}(t))\right]+\int_Vd^3x\frac{1}{c^2}(\boldsymbol{E}\times\boldsymbol{H})_i\\ \\ &=& -A_i(\boldsymbol{z}(t))e+\frac{1}{c^2}\int_Vd^3x(\boldsymbol{E}\times\boldsymbol{H})_i&...&\text{デルタ関数の積分より}\\ \\ &=& \frac{1}{c^2}\int_Vd^3x(\boldsymbol{E}\times\boldsymbol{H})_i-eA_i(\boldsymbol{z}(t))&\\ \\ &\Rightarrow& \frac{1}{c^2}\int_Vd^3x(\boldsymbol{E}\times\boldsymbol{H})-e\boldsymbol{A}(\boldsymbol{z}(t))&...&\text{三次元にした}\\ \\ \end{eqnarray} と書ける。

p.254式(1.19)やp.87式(2.7)など参考。

\(\Box\boldsymbol{A}^c(\boldsymbol{x},t)=-\mu_0\boldsymbol{i}_t(\boldsymbol{x},t)\)と式(2.38)より \begin{eqnarray} &&\Box\boldsymbol{A}^c(\boldsymbol{x},t)-\frac{1}{c^2}\text{grad}\frac{\partial A_0^c(x)}{\partial t}&=&-\mu_0\boldsymbol{i}(x) \\ \\ &\Leftrightarrow& -\mu_0\boldsymbol{i}_t(\boldsymbol{x},t)-\frac{1}{c^2}\text{grad}\frac{\partial A_0^c(x)}{\partial t}&=&-\mu_0\boldsymbol{i}(x) \\ \\ &\Leftrightarrow& \boldsymbol{i}_t(\boldsymbol{x},t)&=&\boldsymbol{i}(x)-\frac{1}{\mu_0c^2}\text{grad}\frac{\partial A_0^c(x)}{\partial t} \\ \\ \end{eqnarray}

\begin{eqnarray} &&-\frac{1}{\mu_0c^2}\text{grad}\frac{\partial}{\partial t}\left(\frac{1}{4\pi\varepsilon_0}\int_V\frac{\rho(\boldsymbol{x}^{\prime},t)}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}d^3x^{\prime}\right)\\ \\ &=& -\frac{1}{\mu_0c^2}\text{grad}\left(\frac{1}{4\pi\varepsilon_0}\int_V\frac{\frac{\partial}{\partial t}\rho(\boldsymbol{x}^{\prime},t)}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}d^3x^{\prime}\right)\\ \\ &=& -\frac{1}{\mu_0c^2}\text{grad}\left(\frac{1}{4\pi\varepsilon_0}\int_V\frac{-\text{div}^{\prime}\boldsymbol{i}(\boldsymbol{x}^{\prime},t)}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}d^3x^{\prime}\right)&...&\text{電荷保存則p.20式(6.1)}\\ \\ &=& \frac{1}{\mu_0c^2}\text{grad}\left(\frac{1}{4\pi\varepsilon_0}\int_V\frac{\text{div}^{\prime}\boldsymbol{i}(\boldsymbol{x}^{\prime},t)}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}d^3x^{\prime}\right)&\\ \\ &=& -\frac{1}{\mu_0c^2}\text{grad}\left(\frac{1}{4\pi\varepsilon_0}\int_V\boldsymbol{i}(\boldsymbol{x}^{\prime},t)\text{grad}^{\prime}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}d^3x^{\prime}\right)&...&\text{(1)部分積分を利用}\\ \\ \end{eqnarray} となる。

1. (1)部分積分の計算
▼ 詳細を開く
\begin{eqnarray} &&\text{div}^{\prime}(\boldsymbol{i}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|})&=&\text{div}^{\prime}\boldsymbol{i}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}+\boldsymbol{i}\text{grad}^{\prime}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}&...&\text{p.445式(A・44)より} \\ \\ &\Leftrightarrow& \int_Vd^3x\text{div}^{\prime}(\boldsymbol{i}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|})&=&\int_Vd^3x\text{div}^{\prime}\boldsymbol{i}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}+\int_Vd^3x\boldsymbol{i}\text{grad}^{\prime}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}&...&\text{両辺を積分した} \\ \\ &\Leftrightarrow& \int_SdS(\boldsymbol{i}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|})\boldsymbol{n}&=&\int_Vd^3x\text{div}^{\prime}\boldsymbol{i}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}+\int_Vd^3x\boldsymbol{i}\text{grad}^{\prime}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}&...&\text{p.445式(A・42)Gaussの定理より} \\ \\ &\Leftrightarrow& 0&=&\int_Vd^3x\text{div}^{\prime}\boldsymbol{i}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}+\int_Vd^3x\boldsymbol{i}\text{grad}^{\prime}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}&...&\text{十分大きな領域の表面では電流は0になるから} \\ \\ &\Leftrightarrow& \int_Vd^3x\text{div}^{\prime}\boldsymbol{i}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}&=&-\int_Vd^3x\boldsymbol{i}\text{grad}^{\prime}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}&\\ \\ \end{eqnarray} となる。

\begin{eqnarray} &&-\frac{1}{\mu_0c^2}\text{grad}\left(\frac{1}{4\pi\varepsilon_0}\int_V\boldsymbol{i}(\boldsymbol{x}^{\prime},t)\text{grad}^{\prime}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}d^3x^{\prime}\right)&\\ \\ &=& -\frac{1}{\mu_0c^2}\frac{1}{4\pi\varepsilon_0}\text{grad}\left(\int_Ve_j\boldsymbol{u}_j(t)\delta^3(\boldsymbol{x}^{\prime}-\boldsymbol{z}_j(t))\text{grad}^{\prime}\frac{1}{|\boldsymbol{x}-\boldsymbol{x}^{\prime}|}d^3x^{\prime}\right)&\\ \\ &=& -\frac{\varepsilon_0\mu_0}{\mu_0}\frac{1}{4\pi\varepsilon_0}\text{grad}e_j\boldsymbol{u}_j(t)\text{grad}^{\prime}\frac{1}{|\boldsymbol{x}-\boldsymbol{z}_j(t)|}&...&\text{デルタ関数の積分で}\boldsymbol{x}^{\prime}=\boldsymbol{z}_j(t)\\ \\ &=& -\frac{1}{4\pi}\text{grad}e_j\boldsymbol{u}_j(t)\text{grad}_{z}\frac{1}{|\boldsymbol{x}-\boldsymbol{z}_j(t)|}&\\ \\ \end{eqnarray} となるので、これを用いて、 \begin{eqnarray} i_{t,\alpha}(\boldsymbol{x},t) &=& i_{\alpha}(\boldsymbol{x},t)-\frac{1}{\mu_0c^2}\text{grad}_{\alpha}\frac{\partial A_0^c(\boldsymbol{x},t)}{\partial t}&...&\text{式(2.60)の\(\alpha\)成分} \\ \\ &=& eu_{\alpha}(t)\delta^3(\boldsymbol{x}-\boldsymbol{z(t)})-\frac{1}{4\pi}\text{grad}_{\alpha}e\boldsymbol{u}_j(t)\text{grad}^{\prime}\frac{1}{|\boldsymbol{x}-\boldsymbol{z}_j(t)|}&\\ \\ &=& eu_{\alpha}(t)\delta^3(\boldsymbol{x}-\boldsymbol{z(t)})-\frac{1}{4\pi}\frac{\partial}{\partial x_{\alpha}}e\boldsymbol{u}(t)\text{grad}^{\prime}\frac{1}{|\boldsymbol{x}-\boldsymbol{z}_(t)|}&\\ \\ &=& e\left[u_{\alpha}(t)\delta^3(\boldsymbol{x}-\boldsymbol{z(t)})-\frac{1}{4\pi}\frac{\partial}{\partial x_{\alpha}}\boldsymbol{u}(t)\text{grad}^{\prime}\frac{1}{|\boldsymbol{x}-\boldsymbol{z}_(t)|}\right]&\\ \\ &=& e\left[u_{\alpha}(t)\delta^3(\boldsymbol{x}-\boldsymbol{z(t)})-\frac{1}{4\pi}\frac{\partial}{\partial x_{\alpha} }(u_{1}(t)\frac{\partial}{\partial z_{1} }\frac{1}{|\boldsymbol{x}-\boldsymbol{z}(t)|}+u_{2}(t)\frac{\partial}{\partial z_2 }\frac{1}{ |\boldsymbol{x}-\boldsymbol{z}(t)|}+u_{3}(t)\frac{\partial}{\partial z_3 }\frac{1}{|\boldsymbol{x}-\boldsymbol{z}(t)| } )\right]&...&\boldsymbol{z}=(z_1,z_2,z_3)\text{としている}\\ \\ &=& e\left[u_{\alpha}(t)\delta^3(\boldsymbol{x}-\boldsymbol{z(t)})-\frac{1}{4\pi}\frac{\partial}{\partial x_{\alpha}}\displaystyle\sum_{\beta}u_{\beta}(t)\frac{\partial}{\partial z_{\beta}}\frac{1}{|\boldsymbol{x}-\boldsymbol{z}(t)|}\right]&\\ \\ &=& e\left[u_{\alpha}(t)\delta^3(\boldsymbol{x}-\boldsymbol{z(t)})-\frac{1}{4\pi}\displaystyle\sum_{\beta}u_{\beta}(t)\frac{\partial}{\partial x_{\alpha}}\frac{\partial}{\partial z_{\beta}}\frac{1}{|\boldsymbol{x}-\boldsymbol{z}(t)|}\right]&\\ \\ \end{eqnarray} となる。

一行目から三行目(二つ目のイコールの先の一項目)へはデルタ関数の積分によって、\(\boldsymbol{x}^{\prime}=\boldsymbol{z}(t)\)としている。
二行目(一つ目のイコール)から三行目の二項目(二つ目のイコールの先の二項目)へは変数の変換を行っている。 \begin{eqnarray} h=x_i^{\prime}-z_i \end{eqnarray} として、 \begin{eqnarray} \frac{\partial}{\partial x_i^{\prime}} &=& \frac{\partial h}{\partial x_i^{\prime}}\frac{\partial }{\partial h} \\ \\ &=& \frac{\partial h}{\partial x_i^{\prime}}\frac{\partial z_i}{\partial h}\frac{\partial }{\partial z_i} \\ \\ &=& 1\frac{\partial z_i}{\partial h}\frac{\partial }{\partial z_i} \\ \\ &=& -1\frac{\partial }{\partial z_i} \\ \\ &=& -\frac{\partial }{\partial z_i} \\ \\ \end{eqnarray} と変換することで符号が逆になっている。

一行目から二行目への計算ではp.85式(2.1)とp.87式(2.6)を用いて、 \begin{eqnarray} \Delta_z\left(\frac{1}{4\pi|\boldsymbol{x}^{\prime}-\boldsymbol{z}(t)|}\right)&=&-\delta^3(\boldsymbol{x}^{\prime}-\boldsymbol{z}(t)) \end{eqnarray} となる式変形を用いている。

一行目から二行目へは \begin{eqnarray} &&\frac{\partial}{\partial z_{\alpha}}\frac{\partial}{\partial z_{\beta}}\frac{|\boldsymbol{x}-\boldsymbol{z}|}{2} \\ \\ &=& -\frac{\partial}{\partial z_{\alpha}}\frac{x_{\beta}-z_{\beta}}{2|\boldsymbol{x}-\boldsymbol{z}|}&...&(1) \\ \\ &=& -\frac{\frac{\partial}{\partial z_{\alpha}}(x_{\beta}-z_{\beta})2|\boldsymbol{x}-\boldsymbol{z}|-(x_{\beta}-z_{\beta})\frac{\partial}{\partial z_{\alpha} }(2|\boldsymbol{x}-\boldsymbol{z}|) }{2|\boldsymbol{x}-\boldsymbol{z}|^2} \\ \\ &=& -\frac{-\delta_{\alpha,\beta}2|\boldsymbol{x}-\boldsymbol{z}|+(x_{\beta}-z_{\beta})\frac{x_{\alpha}-z_{\alpha}}{2|\boldsymbol{x}-\boldsymbol{z}|} }{4|\boldsymbol{x}-\boldsymbol{z}|^2}&...&\text{一項目は}\alpha=\beta\text{の時だけ微分結果が値を持つためKronecherのデルタとした} \\ \\ &=& \frac{\delta_{\alpha,\beta}2|\boldsymbol{x}-\boldsymbol{z}|^2-(x_{\beta}-z_{\beta})\frac{(x_{\alpha}-z_{\alpha})}{2} }{4|\boldsymbol{x}-\boldsymbol{z}|^3} \\ \\ &=& \frac{-(x_{\beta}-z_{\beta})(x_{\alpha}-z_{\alpha}) }{2|\boldsymbol{x}-\boldsymbol{z}|^3}+\frac{\delta_{\alpha,\beta} }{2|\boldsymbol{x}-\boldsymbol{z}|} \\ \\ \end{eqnarray} となる式変形を用いている。

1. (1)積分の計算
▼ 詳細を開く
\(|\boldsymbol{x}-\boldsymbol{z}|\)の微分をする。\(\boldsymbol{x}=(x_1,x_2,x_3),\boldsymbol{z}=(z_1,z_2,z_3)\)とすると \begin{eqnarray} \frac{\partial}{\partial z_1}|\boldsymbol{x}-\boldsymbol{z}| &=& \frac{\partial}{\partial z_1}\sqrt{(x_1-z_1)^2+(x_2-z_2)^2+(x_3-z_3)^2} \\ \\ &=& -\frac{1}{2}\frac{2(x_1-z_1)}{\sqrt{(x_1-z_1)^2+(x_2-z_2)^2+(x_3-z_3)^2}} \\ \\ &=& -\frac{x_1-z_1}{\sqrt{(x_1-z_1)^2+(x_2-z_2)^2+(x_3-z_3)^2}} \\ \\ \end{eqnarray} となることから、一般的に \begin{eqnarray} \frac{\partial}{\partial z_i}|\boldsymbol{x}-\boldsymbol{z}| &=& -\frac{x_i-z_i}{\sqrt{(x_1-z_1)^2+(x_2-z_2)^2+(x_3-z_3)^2}} \\ \\ \end{eqnarray} となる。

\begin{eqnarray} \boldsymbol{A}^c(\boldsymbol{x},t) &=& \displaystyle\sum_{j=1}^N\left( \begin{array}{cccc} A_1^c(\boldsymbol{x},t) \\ A_2^c(\boldsymbol{x},t) \\ A_3^c(\boldsymbol{x},t) \\ \end{array} \right) \\ \\ &=& \displaystyle\sum_{j=1}^N\left( \begin{array}{cccc} \frac{\mu_0 e_j}{4\pi}\left[\frac{u_{1j}(t)}{|\boldsymbol{x}-\boldsymbol{z}_j|}+\displaystyle\sum_{\beta}u_{j\beta}\left\{\frac{(x_1-z_1)(x_{\beta}-z_{\beta})}{2|\boldsymbol{x}-\boldsymbol{z}_j|^3}-\frac{\delta_{1\beta}}{2|\boldsymbol{x}-\boldsymbol{z}_j|}\right\}\right] \\ \\ \frac{\mu_0 e_j}{4\pi}\left[\frac{u_{2j}(t)}{|\boldsymbol{x}-\boldsymbol{z}_j|}+\displaystyle\sum_{\beta}u_{j\beta}\left\{\frac{(x_2-z_2)(x_{\beta}-z_{\beta})}{2|\boldsymbol{x}-\boldsymbol{z}_j|^3}-\frac{\delta_{2\beta}}{2|\boldsymbol{x}-\boldsymbol{z}_j|}\right\}\right] \\ \\ \frac{\mu_0 e_j}{4\pi}\left[\frac{u_{3j}(t)}{|\boldsymbol{x}-\boldsymbol{z}_j|}+\displaystyle\sum_{\beta}u_{j\beta}\left\{\frac{(x_3-z_3)(x_{\beta}-z_{\beta})}{2|\boldsymbol{x}-\boldsymbol{z}_j|^3}-\frac{\delta_{3\beta}}{2|\boldsymbol{x}-\boldsymbol{z}_j|}\right\}\right] \\ \\ \end{array} \right) \\ \\ &=& \displaystyle\sum_{j=1}^N\left( \begin{array}{cccc} \frac{\mu_0 e_j}{4\pi}\left[\frac{u_{1j}(t)}{|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{u_{j1}(x_1-z_1)(x_{1}-z_{1})+u_{j2}(x_1-z_1)(x_{2}-z_{2})+u_{j3}(x_1-z_1)(x_{3}-z_{3})}{2|\boldsymbol{x}-\boldsymbol{z}_j|^3}-\frac{u_{j1}}{2|\boldsymbol{x}-\boldsymbol{z}_j|}\right] \\ \\ \frac{\mu_0 e_j}{4\pi}\left[\frac{u_{2j}(t)}{|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{u_{j1}(x_2-z_2)(x_{1}-z_{1})+u_{j2}(x_2-z_2)(x_{2}-z_{2})+u_{j3}(x_2-z_2)(x_{3}-z_{3})}{2|\boldsymbol{x}-\boldsymbol{z}_j|^3}-\frac{u_{j2}}{2|\boldsymbol{x}-\boldsymbol{z}_j|}\right] \\ \\ \frac{\mu_0 e_j}{4\pi}\left[\frac{u_{3j}(t)}{|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{u_{j1}(x_3-z_3)(x_{1}-z_{1})+u_{j2}(x_3-z_3)(x_{2}-z_{2})+u_{j3}(x_3-z_3)(x_{3}-z_{3})}{2|\boldsymbol{x}-\boldsymbol{z}_j|^3}-\frac{u_{j3}}{2|\boldsymbol{x}-\boldsymbol{z}_j|}\right] \\ \\ \end{array} \right) \\ \\ &=& \displaystyle\sum_{j=1}^N\left( \begin{array}{cccc} \frac{\mu_0 e_j}{4\pi}\left[\frac{u_{1j}(t)}{2|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{u_{j1}(x_1-z_1)(x_{1}-z_{1})+u_{j2}(x_{2}-z_{2})(x_1-z_1)+u_{j3}(x_{3}-z_{3})(x_1-z_1)}{2|\boldsymbol{x}-\boldsymbol{z}_j|^3}\right] \\ \\ \frac{\mu_0 e_j}{4\pi}\left[\frac{u_{2j}(t)}{2|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{u_{j1}(x_{1}-z_{1})(x_2-z_2)+u_{j2}(x_2-z_2)(x_{2}-z_{2})+u_{j3}(x_{3}-z_{3})(x_2-z_2)}{2|\boldsymbol{x}-\boldsymbol{z}_j|^3}\right] \\ \\ \frac{\mu_0 e_j}{4\pi}\left[\frac{u_{3j}(t)}{2|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{u_{j1}(x_{1}-z_{1})(x_3-z_3)+u_{j2}(x_{2}-z_{2})(x_3-z_3)+u_{j3}(x_3-z_3)(x_{3}-z_{3})}{2|\boldsymbol{x}-\boldsymbol{z}_j|^3}\right] \\ \\ \end{array} \right) \\ \\ &=& \displaystyle\sum_{j=1}^N\left( \begin{array}{cccc} \frac{\mu_0 e_j}{4\pi}\left[\frac{u_{1j}(t)}{2|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{\boldsymbol{u}\cdot(\boldsymbol{x}-\boldsymbol{z})(x_1-z_1)}{2|\boldsymbol{x}-\boldsymbol{z}_j|^3}\right] \\ \\ \frac{\mu_0 e_j}{4\pi}\left[\frac{u_{2j}(t)}{2|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{\boldsymbol{u}\cdot(\boldsymbol{x}-\boldsymbol{z})(x_2-z_2)}{2|\boldsymbol{x}-\boldsymbol{z}_j|^3}\right] \\ \\ \frac{\mu_0 e_j}{4\pi}\left[\frac{u_{3j}(t)}{2|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{\boldsymbol{u}\cdot(\boldsymbol{x}-\boldsymbol{z})(x_3-z_3)}{2|\boldsymbol{x}-\boldsymbol{z}_j|^3}\right] \\ \\ \end{array} \right) \\ \\ &=& \displaystyle\sum_{j=1}^N\left( \begin{array}{cccc} \frac{\mu_0 e_j}{8\pi}\left[\frac{u_{1j}(t)}{|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{\boldsymbol{u}\cdot(\boldsymbol{x}-\boldsymbol{z})(x_1-z_1)}{|\boldsymbol{x}-\boldsymbol{z}_j|^3}\right] \\ \\ \frac{\mu_0 e_j}{8\pi}\left[\frac{u_{2j}(t)}{|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{\boldsymbol{u}\cdot(\boldsymbol{x}-\boldsymbol{z})(x_2-z_2)}{|\boldsymbol{x}-\boldsymbol{z}_j|^3}\right] \\ \\ \frac{\mu_0 e_j}{8\pi}\left[\frac{u_{3j}(t)}{|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{\boldsymbol{u}\cdot(\boldsymbol{x}-\boldsymbol{z})(x_3-z_3)}{|\boldsymbol{x}-\boldsymbol{z}_j|^3}\right] \\ \\ \end{array} \right) \\ \\ &=& \displaystyle\sum_{j=1}^N\frac{\mu_0 e_j}{8\pi}\left[\frac{\boldsymbol{u}_{j}(t)}{|\boldsymbol{x}-\boldsymbol{z}_j|}+\frac{\boldsymbol{u}\cdot(\boldsymbol{x}-\boldsymbol{z})}{|\boldsymbol{x}-\boldsymbol{z}_j|^3}(\boldsymbol{x}-\boldsymbol{z})\right] \\ \\ \end{eqnarray} が得られる。